example problem in linear differential equation

$A graph with domain $0 \le t \le 10$ and range $0 \le v(t) \le 90$. There are a series of solutions graphed corresponding to values of c. The graph for c=0 is a horizontal line at v=50. The graphs for c=-20 and c=-40 are below the graph of c=0 and increase up towards the c=0 graph. They shallow out as the near v=50 and don’t cross the horizontal line. The graph of c=-40 is below the graph of c=-20. The graphs for c=20 and c=40 are above the graph of c=0 and decrease down towards the c=0 graph. They shallow out as the near v=50 and don’t cross the horizontal line. The graph of c=40 is above the graph of c=20.$

So, it looks like we did pretty good sketching the graphs back in the direction field section.

Now, recall from the Definitions section that the Initial Condition(s) will allow us to zero in on a particular solution. Solutions to first order differential equations (not just linear as we will see) will have a single unknown constant in them and so we will need exactly one initial condition to find the value of that constant and hence find the solution that we were after. The initial condition for first order differential equations will be of the form

Recall as well that a differential equation along with a sufficient number of initial conditions is called an Initial Value Problem (IVP).

To find the solution to an IVP we must first find the general solution to the differential equation and then use the initial condition to identify the exact solution that we are after. So, since this is the same differential equation as we looked at in Example 1 , we already have its general solution.

Now, to find the solution we are after we need to identify the value of $c$ that will give us the solution we are after. To do this we simply plug in the initial condition which will give us an equation we can solve for $c$. So, let's do this

So, the actual solution to the IVP is.

A graph of this solution can be seen in the figure above.

Let’s do a couple of examples that are a little more involved.

Rewrite the differential equation to get the coefficient of the derivative to be one.

Now find the integrating factor.

Can you do the integral? If not rewrite tangent back into sines and cosines and then use a simple substitution. Note that we could drop the absolute value bars on the secant because of the limits on $x$. In fact, this is the reason for the limits on $x$. Note as well that there are two forms of the answer to this integral. They are equivalent as shown below. Which you use is really a matter of preference.

Also note that we made use of the following fact.

This is an important fact that you should always remember for these problems. We will want to simplify the integrating factor as much as possible in all cases and this fact will help with that simplification.

Now back to the example. Multiply the integrating factor through the differential equation and verify the left side is a product rule. Note as well that we multiply the integrating factor through the rewritten differential equation and NOT the original differential equation. Make sure that you do this. If you multiply the integrating factor through the original differential equation you will get the wrong solution!

Integrate both sides.

Note the use of the trig formula $\sin \left( {2\theta } \right) = 2\sin \theta \cos \theta $ that made the integral easier. Next, solve for the solution.

Finally, apply the initial condition to find the value of $c$.

The solution is then.

Below is a plot of the solution.

$A graph with domain $0 \le x \le 20$ and range $-6 \le y \le 6$. The graph is essentially the graph of a cosine function with an amplitude of 6 and a wavelength of approximately 6.$

First, divide through by the t to get the differential equation into the correct form.

Now let’s get the integrating factor, $\mu \left( t \right)$.

Now, we need to simplify $\mu \left( t \right)$. However, we can’t use $\eqref{eq:eq11}$ yet as that requires a coefficient of one in front of the logarithm. So, recall that

and rewrite the integrating factor in a form that will allow us to simplify it.

We were able to drop the absolute value bars here because we were squaring the $t$, but often they can’t be dropped so be careful with them and don’t drop them unless you know that you can. Often the absolute value bars must remain.

Now, multiply the rewritten differential equation (remember we can’t use the original differential equation here…) by the integrating factor.

Integrate both sides and solve for the solution.

Finally, apply the initial condition to get the value of $c$.

The solution is then,

Here is a plot of the solution.

$A graph with domain $0 \le x \le 20$ and range $0 \le y \le 25$. The starts at approximately (0,25) and decreases sharply until reaching approximately (0.1,1) and then starts to increase for the rest of the graph. The rate of increase is shallow at first then steepens out until it seems to be at approximately a 45 degree angle with the x-axis at the right end of the graph.$

First, divide through by $t$ to get the differential equation in the correct form.

Now that we have done this we can find the integrating factor, $\mu \left( t \right)$.

Do not forget that the “-” is part of $p(t)$. Forgetting this minus sign can take a problem that is very easy to do and turn it into a very difficult, if not impossible problem so be careful!

Now, we just need to simplify this as we did in the previous example.

Again, we can drop the absolute value bars since we are squaring the term.

Now multiply the differential equation by the integrating factor (again, make sure it’s the rewritten one and not the original differential equation).

Apply the initial condition to find the value of $c$.

The solution is then

$A graph with domain $0 \le x \le 7$ and range $0 \le y \le 5000$. The starts at approximately (1,0) has a very shallow increasing slope. Between $3<x<6$ the graph steepens somewhat and starting at x=6 it sharply increases.$

Let’s work one final example that looks more at interpreting a solution rather than finding a solution.

First, divide through by a 2 to get the differential equation in the correct form.

Now find $\mu \left( t \right)$.

Multiply $\mu \left( t \right)$through the differential equation and rewrite the left side as a product rule.

Integrate both sides (the right side requires integration by parts – you can do that right?) and solve for the solution.

Apply the initial condition to find the value of $c$ and note that it will contain $y_{0}$ as we don’t have a value for that.

So, the solution is

Now that we have the solution, let’s look at the long term behavior ( i.e. $t \to \infty $) of the solution. The first two terms of the solution will remain finite for all values of $t$. It is the last term that will determine the behavior of the solution. The exponential will always go to infinity as $t \to \infty $, however depending on the sign of the coefficient $c$ (yes we’ve already found it, but for ease of this discussion we’ll continue to call it $c$). The following table gives the long term behavior of the solution for all values of $c$.

This behavior can also be seen in the following graph of several of the solutions.

$A graph with domain $0 \le t \le 8$ and range $-8 \le y \le 8$. There are a series of graph representing different values of c on the graph. The c=0 graph is essentially an oscillation along the t-axis. Graphs corresponding to c>0 (there are 5 of them with no actual c values indicated) all start with y>0 and increase and have an oscillation to them as the increase. The farther up the y axis they start the faster they increase. Graphs corresponding to c<0 (there are 5 of them with no actual c values indicated) all start with y<0 and decrease and have an oscillation to them as the decrease. The farther down the y axis they start the faster they decrease.$

Now, because we know how $c$ relates to $y_{0}$ we can relate the behavior of the solution to $y_{0}$. The following table give the behavior of the solution in terms of $y_{0}$ instead of $c$.

Note that for ${y_0} = - \frac{{24}}{{37}}$ the solution will remain finite. That will not always happen.

Investigating the long term behavior of solutions is sometimes more important than the solution itself. Suppose that the solution above gave the temperature in a bar of metal. In this case we would want the solution(s) that remains finite in the long term. With this investigation we would now have the value of the initial condition that will give us that solution and more importantly values of the initial condition that we would need to avoid so that we didn’t melt the bar.

Linear Differential Equation

Linear differential equation is an equation having a variable, a derivative of this variable, and a few other functions. The standard form of a linear differential equation is dy/dx + Py = Q, and it contains the variable y, and its derivatives. The P and Q in this differential equation are either numeric constants or functions of x.

The linear differential equation in an important form of a differential equation and can be solved using a formula. Let us learn the formula and derivation, to find the general solution of a linear differential equation.

What Is a Linear Differential Equation?

The linear differential equation is of the form dy/dx + Py = Q, where P and Q are numeric constants or functions in x. It consists of a y and a derivative of y. The differential is a first-order differentiation and is called the first-order linear differential equation.

This linear differential equation is in y. Similarly, we can write the linear differential equation in x also. The linear differential equation in x is dx/dy + $P_1$x = $Q_1$.

Linear Differential Equations in x and y

Some of the examples of linear differential equation in y are dy/dx + y = Cosx, dy/dx + (-2y)/x = x 2 .e -x . And the examples of linear differential equation in x are dx/dy + x = Siny, dx/dy + x/y = ey. dx/dy + x/(ylogy) = 1/y.

Derivation for Solution of Linear Differential Equation

The derivation for the general solution for the linear differential equation can be understood through the below sequence of steps. The first-order differential equation is of the form.

dy/dx +Px = Q

Here we multiply both sides of the equation by a function of x, say g(x) . Further, this function is chosen such that the right hand side of the equation is derivative of y.g(x). d/dx(y.g(x)) = y.g(x).

g(x).dy/dx + P.g(x).y = Q.g(x)

Choose g(x) in such a way such that the RHS becomes the derivative of y.g(x).

g(x).dy/dx + P.g(x)y = d/dx(y.g(x)]

The right hand side of the above expression is derived using the derivative formula for the product of functions.

g(x).dy/dx + P.g(x).y = g(x).dy/dx + y.g'(x)

P.g(x) = g'(x)

P = g'(x)/g(x)

Integrating both sides with respect to x, we get

$\int P.dx= \int \frac{g'(x)}{g(x)}.dx$

$\int P.dx= log(g(x))$

$g(x)= e^{\int P.dx}$

This function $g(x)= e^{\int P.dx}$ is called the Integrating Factor (I.F) of the given linear differential equation. Substituting the value of g(x) in equation linear differential equation, the following expression is obtained.

$e^{\int P.dx}.\dfrac{dy}{dx} + Pe^{\int P.dx}y = Q.e^{\int P.dx}$

$\dfrac{d}{dx}(y.e^{\int P.dx} )= Qe^{\int P.dx}$

Integrating both sides, with respect to x the following expression is obtained..

$y.e^{\int P.dx} =\int (Q.e^{\int P.dx}.dx)$

$y=e^{-\int P.dx} .\int (Q.e^{\int P.dx}.dx) + C$

The above expression is the general solution of the linear differential eqution.

dy/dx - y/x = 2x

Comparing this with the differential equation dy/dx + Py = Q we have the values of P = -1/x and the value of Q = 2x. Hence we have the integration factor as IF = $e^{\int -\dfrac{1}{x}.dx}$ = $e^{-\log x}$ = $\frac{1}{x}$.

Further, the solution of the differential equation is as follows.

y$\frac{1}{x}$ = $\int 2x.\frac{1}{y} .dx + c$ $\frac{y}{x}$ = $\int 2.dx + c$ $\frac{y}{x}$ = 2x + c y = 2x 2 + xc Answer: Thus the general solution of the given linear differential equation is y = 2x 2 + xc

Example 2: Find the derivative of dy/dx + Secx.y = Tanx

The given differential equation is dy/dx + Secx.y = Tanx. Comparing this with the linear differential equation dy/dx + Px = Q, we have P = Secx, and Q = Tanx.

Further the integrating factor is I.F = \(e^{\int Secx.dx}

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example problem in linear differential equation

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Practice Questions on Linear Differential Equation

Faqs on linear differential equation.

The linear differential equation is an equation having a variable, a derivative of this variable, and a few other functions. The standard form of a linear differential equation is dy/dx + Py = Q, and it contains the variable y, and its derivatives. The P and Q in this differential equation are either numeric constants or functions of x.

How Do You Know If a Differential Equation Is a Linear Differential Equation?

A differential equation is said to be a linear differential equation if it has a variable and its first derivative. The linear differential equation in y is of the form dy/dx + Py = Q, Here we have the variable y, the first derivative of the variable y, and we have P, Q which are functions in x. From the name of linear, these differential equations have only the first degree derivatives.

How Do You Solve a Linear Differential Equation?

The solution of a linear differential equation is through three simple steps. First simplify and write the given differential equation in the form dy/dx + Py = Q. For this find the Integrating Factor (IF) = $e^{\int P.dx}$. Finally the solution of the linear differential equation is $y(I.F) = \int(Q × I.F).dx + C$

What Is the Standard Form of Linear Differential Equation in x?

The standard form of the linear differential equation in x is dx/dy + Px = Q, This is a differential equation having a variable x, the first derivative of x, and P, Q represent the functions in y. The linear differential equation in x has first-order derivative of x.

What Is the Formula For the General Solution of Linear Differential Equation

The formula for general solution of the differential equation dy/x +Py = Q is $y.(I.F)=\int (Q.(I.F).dx)+ C$. Here we have Integrating Factor (I.F) = $e^{\int P.dx}$. Also the formula for the general solution of the differential equation dx/y +Px = Q is $x.(I.F)=\int (Q.(I.F).dy)+ C$. Here we have Integrating Factor (I.F) = $e^{\int P.dy}$.

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Linear vs nonlinear differential equation

How to distinguish linear differential equations from nonlinear ones? I know, that e.g.: $$ y''-2y = \ln(x) $$ is linear, but $$ 3+ yy'= x - y $$ is nonlinear. Why?

ordinary-differential-equations

5 Answers 5

Linear differential equations are those which can be reduced to the form $Ly = f$, where $L$ is some linear operator.

Your first case is indeed linear, since it can be written as:

$$\left(\frac{d^2}{dx^2} - 2\right)y = \ln(x)$$

While the second one is not. To see this first we regroup all $y$ to one side:

$$y(y'+1) = x - 3$$

then we simply notice that the operator $y\mapsto g(y) = y(y'+1)$ is not linear (for example we can take two functions $y_1$ and $y_2$ and notice that $g(y_1+y_2)\neq g(y_1) + g(y_2)$).

If the equation would have had $\ln (y)$ on the right, that also would have made it non-linear, since natural logs are non-linear functions. Remember that this has its roots in linear algebra: $y=mx+b$. You can analyse functions term-by-term to determine if they are linear, if that helps. The first time a term is non-linear, then the entire equation is non-linear.

Remember that the $x$s can pretty much do or appear however they want, since they're independent. Which means if you can't tell just by glancing, try to group all your $y$ terms to one side and then analyse them. Makes it much easier.

See, I was also overthinking this, but realised you have to go back to those definitions we're given.

Two criteria for linearity:

The dependent variable y and its derivatives are of first degree; the power of each y is 1. $\frac{dy}{dx}$; yes. $(\frac{dy}{dx})^4$, no.

Each coefficient depends only on the independent variable $x$.

$yy'$ makes it nonlinear as has been said, because that coefficient on $y'$ is not in $x$. Had that coefficient been a constant, you would have been correct to call it linear, since constants can be functions of $x$. Like, $f(3)=x$. Its graph is a line, i.e. linear function.

Always go back to the definitions. :-)

1 $\begingroup$ I have edited your answer for better readability. For some basic information about writing math at this site see e.g. here , here , here and here . $\endgroup$ – Jesse P Francis Nov 13, 2015 at 4:04

One could define a linear differential equation as one in which linear combinations of its solutions are also solutions.

Linear Differential equations are those in which the dependent variable and its derivatives appear only in first degree and not multiplied together

7 $\begingroup$ $y'''+y''+y=e^x$ is linear ;) The degree is irrelevant. What's important is "not multiplied together" $\endgroup$ – Scientifica Aug 28, 2016 at 6:21

Because highest order derivative is multiplied with dependent variable $y$. Like $y y'$.

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Linear Differential Equation

A differential equation (an equation that connects one or more functions and their derivatives) interpreted through a linear expression that has 1 or more terms being algebraic in the function which is unknown and its derivatives is a linear differential equation in mathematics. It is of the numerical form

Aforesaid equations are ordinary differential equations. If the function that is unknown is dependent on various variables & the derivatives present in the equation are derivatives that are partial in nature, then a linear differential equation is a linear partial differential equation. A linear differential equation or system of equations that are linear in a way that the related homogeneous equations consist of coefficients that are constant can be figured out by the method of quadrature, which implies the solutions can be represented in form of integrals. This holds good for the linear equation with order 1, with coefficients that are non-constant. A linear combination consisting of rudimentary differential operators, with coefficients as differentiable functions. The linear operator in the univariate scenario is given by

Linear Differential Equation Formula

The standard form of a linear differential equation is (dy / dx) + Py = Q. Here P and Q are constants in x. It possesses the term y and its derivative. It is of first-order and hence termed first-order linear differential equation. The differential is in terms of y, similarly, it can be written in terms of x also. The linear differential equation in terms of x can be expressed as (dx/dy)+P_{1}x=Q_{1}

Linear Differential Equation Properties

The linear differential equations have the following properties.

a] The y function and its respective derivatives come in the equation till the first degree only.

b] The products of y and/or any of its respective derivatives are not present.

c] No functions that are transcendental.

Linear Differential Equation Examples

Example 1: Solve the equation: x \frac{d y}{d x}-2 y=x^{3} \cos 4 x .

Answer:

Convert the given equation into the standard form (dy / dx) + Py = Q of the linear differential equation.

Obtain the values of P and Q by comparing it to the standard form of linear differential equation.

P = (- 2 / x) and Q = x^{2} \cos 4 x

Compute the integrating factor by using the below formula.

IF = η(𝑥) = e^{\int P(x)dx}

The general solution is y =\frac{\int \eta(x) Q(x) d x-C}{\eta(x)} \\ y =\frac{\int x^{-2} x^{2} \cos 4 x d x-C}{x^{-2}} \\ y =x^{2}\left(\frac{\sin 4 x}{4}-C\right)

Example 2: Solve the equation: y’ (x) + y (x) / x = 3x.

Example 3: Solve the equation: y^{\prime}-y-x e^{x}=0

Example 4: Solve the differential equation xy'=y+2x^{3}

This problem can be solved by using the method of variation of a constant. First, find the general solution of the homogeneous equation:

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Differential equations

First order differential equations, second order linear equations, laplace transform.

with Answer, Solution, Formula - Homogeneous Differential Equations: Solved Example Problems | 12th Business Maths and Statistics : Chapter 4 : Differential Equations

Chapter: 12th business maths and statistics : chapter 4 : differential equations, homogeneous differential equations: solved example problems.

Example 4.15

Solve the differential equation y 2 dx + ( xy + x 2 ) dy = 0

Example 4.17

Find the particular solution of the differential equation x 2 dy + y ( x + y ) dx = 0 given that x = 1, y = 1

Example 4.18

If the marginal cost of producing x shoes is given by (3 xy + y 2 ) dx + ( x 2 + xy ) dy = 0 and the total cost of producing a pair of shoes is given by ₹12. Then find the total cost function.

Given marginal cost function is (x 2 + xy) dy + (3xy + y 2 )dx=0

Example 4.19

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Initial Value Problem Differential Equations

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What do the phrases "once upon a time", "in the beginning", and "on a dark and stormy night" all have in common? They all indicate the start of a story! That is what the initial value for a differential equation tells you, where to start. But unlike the fact that two stories that start with "once upon a time" aren't the same, all solutions to initial value problems might not be so unique. So read through to figure out how to make sure you get a unique solution to your problem!

Definition of an Initial-Value Problem

Let's start by answering a common question. Is it supposed to be "initial-value" or "initial value"? Well, it depends on what country you are in and who wrote the textbook you are reading! To be consistent this article doesn't use the hyphen. Instead initial value problem is abbreviated IVP , which gets around the question of whether that hyphen should be there or not.

So what is an IVP?

An IVP is a differential equation together with a place for a solution to start. They are often written

\[ \begin{align} &y' = f(x,y) \\ &y(a) = b \end{align} \]

where $(a,b)$ is the point the solution $y(x)$ must go through.

Remember that the first step to solving a differential equation is to try and find the general solution, which gives you a constant of integration. Then using the initial value, you can find a particular solution. For more information on these topics see General Solutions to Differential Equations and Particular Solutions to Differential Equations.

First Order Differential Equation Initial Value Problem

Let's start with the constant coefficient first order linear differential equation

\[ y'+Ay=B\]

where $A$ and $B$ are real numbers with $A \neq 0$. Remember that the general solution to this linear differential equation is

\[y(x) =Ce^{-Ax}+\frac{B}{A}\]

where $C$ is a constant.

If you add the initial value $y(x_1) = y_1$ where $x_1$ and $y_1$ are real numbers, then you can plug those into the general solution to get

\[ y_1=Ce^{-Ax_1}+\frac{B}{A}, \]

\[ C = \frac{y_1 - \frac{B}{A}}{e^{-Ax_1}}. \]

That means the solution to the IVP

\[ \begin{align} y' &= f(x,y) \\ y(x_1) &= y_1 \end{align} \]

\[ \begin{align} y(x) &= \frac{y_1 - \frac{B}{A}}{e^{-Ax_1}} e^{-Ax}+\frac{B}{A} \\ &= \left(y_1 - \frac{B}{A} \right)e^{-A(x-x_1)} +\frac{B}{A} . \end{align}\]

That can be a bit of a pain to memorize, so rather than doing that it is best to remember the general solution and then solve given your initial values. Let's take a look at an example.

Solve the IVP

\[ \begin{align} &y' +3y = 7 \\ &y(0) = \frac{1}{3}. \end{align} \]

The first step is finding the general solution, which for this problem is

\[y(x) =Ce^{-3x}+\frac{7}{3}.\]

Remember that this is actually a family of functions, and you can graph them for various values of $C$.

The IVP is asking you to pick out the function that goes through the point $\left(0, \frac{1}{3}\right)$. Using the initial value to solve for $C$, you get

\[\frac{1}{3} =Ce^{-3\cdot 0}+\frac{7}{3},\]

so $C = -2$. That means the solution to the IVP is

\[y(x) = -2e^{-3x}+\frac{7}{3}.\]

It is one particular function out of the family of functions that has the property that it goes through the initial value, as you can see in the graph below.

Solutions to linear constant coefficient first order differential equation IVPs have some nice properties that all come directly from the fact that you can explicitly write down the solution as

\[ y(x) = \left(y_1 - \frac{B}{A} \right)e^{-A(x-x_1)} +\frac{B}{A}.\]

Those properties are,

Solutions to the IVP are unique . This means that for every initial value you get one, and only one, value of $C$ that works.

If you have two different initial values, the solutions corresponding to those initial values cannot cross. If they could cross it would violate the uniqueness part.

Solutions exist on the whole real line.

Solutions all have the same long term behavior. In other words, for a solution $y(x)$ of the IVP, \[ \lim\limits_{x \to \infty} y(x) =\begin{cases} \frac{B}{A} & \mbox{ if } A>0 \\ \infty & \mbox{ if } A<0\quad \mbox{provided that } y_1-\frac{B}{A}>0 \\ -\infty & \mbox{ if } A<0\quad \mbox{provided that } y_1-\frac{B}{A}<0\end{cases}. \]

Now let's move on to the first order linear differential equation

\[ y'+P(x)y=Q(x)\]

where $P(x)$ and $Q(x)$ are functions. To find the solution to this differential equation you use the integrating factor

\[ \alpha(x)=e^{\int P(x)\,\mathrm{d}x},\]

and the general solution is

\[ y(x) = \frac{1}{\alpha (x)} \left( \int \alpha(x) \, Q(x) \, \mathrm{d}x + C \right).\]

Finding the solution to the IVP

\[ \begin{align} y'+P(x)y&=Q(x) \\ y(a) &= b \end{align} \]

isn't going to be nearly as nice as in the constant coefficient case since you need to do the integration before you can figure out what $C$ is. So let's look at an example.

\[ \begin{align} &y'+\frac{1}{x}y=\frac{1}{\sqrt{x}} \\ &y(1) = \frac{5}{3} .\end{align} \]

First let's find the general solution. The integrating factor is

\[ \begin{align} \alpha(x)&=e^{\int P(x)\,\mathrm{d}x} \\ &= \exp\left( \int \frac{1}{x} \,\mathrm{d}x \right) \\ &= |x|. \end{align} \]

Notice that the initial value has $x=1$, so in other words you are interested in positive values for $x$ and you can drop the absolute value in the integrating factor to write

\[ \alpha(x) = x.\]

Then the general solution is given by

\[ \begin{align} y(x) &= \frac{1}{\alpha (x)} \left( \int \alpha(x) \, Q(x) \, \mathrm{d}x + C \right) \\ &= \frac{1}{x} \left( \int x \frac{1}{\sqrt{x}} \, \mathrm{d}x + C \right) \\ &= \frac{1}{x} \left( \int \sqrt{x} \, \mathrm{d}x + C \right) \\ &= \frac{1}{x} \left( \frac{2}{3}x^{\frac{3}{2} }+ C \right) \\ &= \frac{2\sqrt{x}}{3} + \frac{C}{x}. \end{align}\]

Now you can plug in the initial value $ y(1) = \dfrac{5}{3} $ to get

\[ \frac{5}{3} = \frac{2\sqrt{1}}{3} + \frac{C}{1}, \]

\[C = \frac{5}{3} - \frac{2}{3} = 1.\]

With the assumption that $x>0$, the solution to the IVP is

\[ y(x) = \frac{2\sqrt{x}}{3} + \frac{1}{x}.\]

Notice that there was only one solution, but it certainly doesn't exist on the whole real line! For examples of a first order linear IVP without a solution, or with infinitely many solutions, take a look at Particular Solutions to Differential Equations .

You might wonder when a first order linear IVP will have a unique solution that exists on the whole real line.

If $a, b \in \mathbb{R}$, and $P(x)$, $Q(x)$ are both continuous functions on the whole real line then the solution to the initial value problem

\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]

exists and is unique on the whole real line .

Initial value Problems and Separable Differential Equations

Remember that a differential equation is separable if you can write it in the form

\[N(y)y' = M(x)\]

where $N(y)$ and $M(x)$ are functions. For more information and examples of this kind of equation see Separable Equations .

To make this into an IVP, all you need to do is pick an initial value. So for real numbers $a$ and $b$, the IVP is

\[\begin{align} &N(y)y' = M(x) \\ &y(a) = b. \end{align}\]

With separable equations, you often need to be careful with the interval of existence for solutions. In cases like that, the initial value tells you which of the intervals to choose as the interval of existence.

Let's take a look at an example.

If possible, solve the IVP

\[ \begin{align} &y' = \frac{y^2}{x} \\ &y(0) = 4. \end{align} \]

First, you can rewrite the differential equation as

\[ \frac{1}{y^2} y' = \frac{1}{x}, \]

so it is a separable differential equation. Separating variables and integrating gives you

\[ \int \frac{1}{y^2} \, \mathrm{d}y = \int \frac{1}{x} \, \mathrm{d}x\]

\[ -\frac{1}{y} = \ln |x| + C.\]

Now let's try and use the initial conditions. If you do, you will be using $x=0$, and you can't take the natural log of zero. So in fact this IVP has no solution.

One more example, to see the kinds of things that can happen.

Consider the IVP

\[ \begin{align} &y' = y^{\frac{1}{3}} \\ &y(0) = 0. \end{align} \]

First show that the constant solution $y(x)=0$ satisfies this IVP. Then see if there are any other solutions.

Certainly if you take the derivative of a constant function you get zero, and $0^\frac{1}{3} = 0$, so the constant function $y(x) = 0$ satisfies the differential equation. It also satisfies the initial condition, so it does satisfy the IVP.

What about other solutions? This is a separable equation, so separating and integrating gives you

\[ \int \frac{1}{y^\frac{1}{3} } \, \mathrm{d}y = \int 1\, \mathrm{d}x\]

\[ \frac{3}{2}y^\frac{2}{3} = x+ C.\]

Using the initial value $y(0) = 0$ to find $C$, you get

\[ \frac{3}{2}0^\frac{2}{3} = 0+ C,\]

so $C=0$. That means there is a second solution to this IVP, namely the implicit solution

\[ \frac{3}{2}y^\frac{2}{3} = x.\]

You can get the explicit solution by solving for $y$. If you do, you get

\[ y^\frac{2}{3} = \frac{2}{3}x,\]

\[ y(x) =\left( \frac{2}{3}x \right)^{\frac{3}{2}}.\]

Notice that the maximal interval of existence is $ [0,\infty) $ since you can't take the square root of a negative number. If you graph the two functions, you can see that both the constant function and the function $y(x)$ that you solved for both satisfy the equation and the initial value.

More examples are always good!

Examples of Differential Equation Initial Value Problems

Let's look more examples involving IVPs.

Find the solution to the IVP

\[ \begin{align} &2xy'+4y=3 \\ &y(2) = \frac{5}{4} \end{align} \]

by first showing that the general solution is

\[y(x) = \frac{C}{x^2} + \frac{3}{4}.\]

What is the interval of existence for the solution to the IVP?

To see that $y(x)$ is a general solution, you will need to find the derivative and plug it into the equation. The derivative is

\[ y'(x) = \frac{-2C}{x^3} .\]

Plugging that in, you get

\[ \begin{align} 2xy'+4y &= 2x\left(\frac{-2C}{x^3} \right) + 4\left(\frac{C}{x^2} + \frac{3}{4} \right) \\ &= \frac{-4C}{x^2} + \frac{4C}{x^2} + 4\left(\frac{3}{4} \right) \\ &= 3, \end{align}\]

which is the right hand side of the original differential equation. Therefore $y(x)$ is a general solution.

Notice that the general solution is not defined at $x = 0$. Therefore the interval of existence is either going to be $ (-\infty , 0)$ or $(0, \infty)$. Since the initial value is when $x=2$, the interval of existence for the IVP is $(0, \infty)$.

Now to solve the IVP. Using the initial value,

\[ y(2) = \frac{C}{2^2} + \frac{3}{4} = \frac{5}{4}. \]

Solving for $C$, you can see that $C=2$. So the solution to the IVP is

\[y(x) = \frac{2}{x^2} + \frac{3}{4}.\]

Sometimes you will have a solution which is implicit, and that can be used to solve an IVP as well.

\[ y^2 = x^2 - 3\]

an implicit solution to the IVP

\[ \begin{align} &y' = \frac{x}{y} \\ &y(2) = 1 ?\end{align} \]

This requires the use of implicit differentiation. But before doing that, it is a good idea to make sure the proposed solution satisfies the initial value, because if it doesn't then it certainly can't be a solution to the IVP! Checking, the left hand side gives you $(y(2))^2 = 1^2 = 1$, and the right hand side gives you $2^2 - 3 = 1$. Since they are the same, the proposed implicit solution at least satisfies the initial value.

Then using implicit differentiation and remembering that $y$ is a function of $x$,

\[ \frac{\mathrm{d}}{\mathrm{d} x} (y(x))^2 = \frac{\mathrm{d}}{\mathrm{d} x} \left(x^2 - 3 \right).\]

Looking first at the left hand side, and using the fact that if it is a solution then

\[y' = \frac{x}{y} ,\]

\[ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x} (y(x))^2 &=2y(x)y'(x) \\ &= 2y\frac{x}{y} \\ &= 2x.\end{align}\]

On the right hand side,

\[ \frac{\mathrm{d}}{\mathrm{d} x} \left(x^2 - 3 \right) = 2x.\]

Since both sides are the same, $y(x)$ does satisfy the differential equation.

Therefore the proposed implicit solution does actually solve the IVP.

What do you do if you can't find an implicit or explicit solution to an IVP?

Numerical Initial Value Problems in Ordinary Differential Equations

Most IVPs you have seen here you could find an explicit solution for, or at least an implicit one! What do you do if you have an IVP that you can't solve like that? That is when you need to use numerical methods to find solutions to IVPs. The method that most people start with is Euler's Method . It uses an initial value, and the differential equation, to calculate slopes that lead to an approximation of the solution to an ordinary differential equation IVP. For more information on this topic, along with plenty of examples, see Euler's Method.

Show question

In terms of the solution of a differential equation, what does the initial value tell you?

It gives you a point the solution must go through.

What is the notation for an IVP?

\[ \begin{align} y' &= f(x,y) \\ y(a) &= b \end{align} \]

What does a first order linear constant coefficient IVP look like?

\[ \begin{align} y'+Ay&=B \\ y(x_1) &= y_1, \end{align}\]

where $A, B, x_1$, and $y_1$ are real numbers with $A \neq 0$.

What is the solution to

\[ \begin{align} y(x) &= \frac{y_1 - \frac{B}{A}}{e^{-Ax_1}} e^{-Ax}+\frac{B}{A} \\ &= \left(y_1 - \frac{B}{A} \right)e^{-A(x-x_1)} +\frac{B}{A} . \end{align}\]

Name 4 properties of first order linear constant coefficient IVPs.

Solutions to the IVP are unique .

If you have two different initial values, the solutions corresponding to those initial values cannot cross.

Solutions all have the same long term behavior.

For what kind of differential equation are you guaranteed a unique solution that exists on the whole real line?

First order linear constant coefficient initial value problems.

When does a first order linear IVP have a unique solution on the whole real line?

If $a, b \in \mathbb{R}$, and $P(x)$, $Q(x)$ are both continuous functions on the whole real line then the solution to the initial value problem

\[\begin{align} y' + P(x)y &= Q(x) \\ y(a) &= b \end{align}\]

What does a separable differential equation IVP look like?

So for real numbers $a$ and $b$, the IVP is

\[\begin{align} N(y)y' &= M(x) \\ y(a) &= b. \end{align}\]

What do you have to look out for when solving separable IVPs?

You need to be sure to pick the interval of existence that contains the initial value.

What can you do if you can't find an explicit or implicit solution to an IVP?

Try a numerical method, like Euler's Method.

What is Euler's Method used for?

Finding a numerical approximation to the solution of an IVP.

Do all IVPs have a solution?

No. Some don't have solutions at all.

Do IVPs all have unique solutions?

No, some IVPs have two, or sometimes even more, solutions!

Will you always be able to find an explicit solution to an IVP, assuming there is a solution?

No. Sometimes all you will be able to find is an implicit solution.

of the users don't pass the Initial Value Problem Differential Equations quiz! Will you pass the quiz?

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Introduction to Differential Equations

Calculus - introduction to differential equations and solved problems - outline of contents:, target audience: high school students, college freshmen and sophomores, students preparing for the international baccalaureate (ib), ap calculus ab, ap calculus bc, a level, singapore/gce a-level; class 11/12 students in india preparing for isc/cbse and entrance examinations like the iit-jee/aieee anyone else who needs this tutorial as a reference, diﬀerential equations, here's a quick outline of what this tutorial introduces :, • diﬀerential equation: , • ordinary diﬀerential equation: , • partial diﬀerential equation: , • order and degree of a diﬀerential equation: , • linear and non-linear diﬀerential equations: , • general,particular and singular solutions: , • initial-value problem(ivp) and boundary-value problem(bvp): , • linear independence and dependence: , • homogeneous equations: , • first order linear diﬀerential equations, • characteristic/auxiliary equation: , in case you'd like to take a look at some of our other tutorials related to single variable calculus :, our calculus tutorials .

Math Insight

Examples of solving linear ordinary differential equations using an integrating factor.

Solve the ODE \begin{gather} \diff{x}{t} -\cos(t)x(t)=\cos(t) \label{example1} \end{gather} for the initial conditions $x(0)=0$.

Solution : Since this is a first order linear ODE, we can solve it by finding an integrating factor $\mu(t)$. If we choose $\mu(t)$ to be \begin{align*} \mu(t) = e^{-\int \cos(t)} = e^{-\sin(t)}, \end{align*} and multiply both sides of the ODE by $\mu$, we can rewrite the ODE as \begin{align*} \diff{}{t}(e^{-\sin(t)}x(t)) &= e^{-\sin(t)}\cos(t). \end{align*} Integrating with respect to $t$, we obtain \begin{align*} e^{-\sin(t)}x(t) &= \int e^{-\sin(t)}\cos(t) dt +C\\ &= -e^{-\sin(t)} + C, \end{align*} where we used the $u$-subtitution $u=\sin(t)$ to compute the integral. Dividing through by $e^{-\sin(t)}$, we calculate that the the general form of the solution to equation \eqref{example1} is \begin{gather*} x(t) = -1 + Ce^{\sin(t)}. \end{gather*}

We verify that we have a solution to equation \eqref{example1}. Since $$\diff{x}{t} = Ce^{\sin(t)}\cos(t)$$ we calculate that $$\diff{x}{t} - \cos(t)x(t) = Ce^{\sin(t)}\cos(t)-\cos(t)(-1 + Ce^{\sin(t)}) = \cos(t),$$ demonstrating that we have found the general solution to the ODE.

We determine the integration constant $C$ by the condition $0=x(0)=-1+Ce^0 = -1+C$, so that $C=1$. Our specific solution to the ODE of \eqref{example1} is \begin{gather*} x(t) = -1 + e^{\sin(t)}. \end{gather*}

Solve the ODE \begin{gather} \diff{x}{t} = \frac{1}{\tau}(-x + I(t)) \label{example2} \end{gather} with initial condition $x(t_0)=x_0$.

Solution : Rewrite the equation in the form \begin{gather*} \diff{x}{t} + \frac{x}{\tau}= \frac{I(t)}{\tau}. \end{gather*} In this case, our integrating factor is $\mu(t) = e^{\int (1/\tau) dt} = e^{t/\tau}$. Multiplying through by $\mu(t)$, we can rewrite our ODE as \begin{gather*} \diff{}{t}\left(e^{t/\tau}x(t)\right)= \frac{I(t)}{\tau}e^{t/\tau}. \end{gather*}

For this example, let's integrate from $t_0$ to $t$, rather than calculate the indefinite integral as in previous examples. \begin{align*} \int_{t_0}^t \diff{}{t'}\left(e^{t'/\tau}x(t')\right)dt' &= \int_{t_0}^t \frac{I(t')}{\tau}e^{t'/\tau}dt'\\ e^{t/\tau}x(t) - e^{t_0/\tau}x(t_0) &= \frac{1}{\tau}\int_{t_0}^t e^{t'/\tau}I(t')dt' \end{align*} Dividing through by $e^{t/\tau}$ and using the initial conditions $x(t_0)=x_0$, the solution to the ODE of \eqref{example2} is \begin{gather} x(t) = e^{-(t-t_0)/\tau}x_0 + \frac{1}{\tau}\int_{t_0}^t e^{-(t-t')/\tau}I(t')dt'. \label{example2sol} \end{gather}

To verify this solution, we differentiate equation \eqref{example2sol} with respect to $t$, obtaining three terms (two from the exponentials and one from the upper integration limit), \begin{align*} \diff{x}{t} &= -\frac{1}{\tau} e^{-(t-t_0)/\tau}x_0 -\frac{1}{\tau} \frac{1}{\tau}\int_{t_0}^t e^{-(t-t')/\tau}I(t')dt' + \frac{1}{\tau} e^{-0/\tau}I(t) \\ &= -\frac{1}{\tau} \left(e^{-(t-t_0)/\tau}x_0 +\frac{1}{\tau}\int_{t_0}^t e^{-(t-t')/\tau}I(t')dt'\right) + \frac{1}{\tau} I(t) \\ &=-\frac{1}{\tau}x(t) + \frac{1}{\tau}I(t). \end{align*} Indeed $x(t)$ satisfies equation \eqref{example2}. If we plug $t=t_0$ into equation \eqref{example2sol}, the integral is from $t_0$ to $t_0$, so is zero. The exponential of the first term is $e^0=1$, and we confirm that $x(t_0)=x_0$.

Solve the ODE \begin{align} \diff{x}{t} + e^{t}x(t) = t^2 \cos(t) \label{example3} \end{align} with the initial condition $x(0)=5$.

Solution : The first step is to find the integrating factor \begin{gather*} \mu(t) = e^{\int e^t dt} = e^{e^t} = \exp(e^t), \end{gather*} where $\exp(x)$ is another way of writing $e^x$. Multiplying equation \eqref{example3} by $\mu(t)$, then the left hand side is the derivative of $\mu(t)x(t)$. We can write it as \begin{align*} \diff{}{t}\left( \exp(e^t) x(t)\right) = t^2 \cos(t) \exp(e^t). \end{align*}

To solve the ODE in terms of the initial conditions $x(0)$, we integrate from $0$ to $t$, obtaining \begin{align*} \int_0^t \diff{}{s}\left( \exp(e^{s}) x(s)\right)ds &= \int_0^t {s}^2 \cos(s) \exp(e^{s})ds\\ \exp(e^{t}) x(t) - \exp(e^{0}) x(0) &= \int_0^t {s}^2 \cos(s) \exp(e^{s})ds. \end{align*} Even though we cannot compute the integral analytically, we still consider the ODE solved. Plugging in the initial conditions $x(0)=5$, we can write the solution of the ODE \eqref{example3} as \begin{align*} x(t) &= 5\exp(1-e^t)+ \int_0^t {s}^2 \cos(s) \exp(e^{s}-e^{t})ds. \end{align*}

You can easily check that $x(t)$ satisfies the ODE \eqref{example3} and the initial conditions $x(0)=5$.

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IMAGES

Linear Differential Equation: Properties, Solving Methods, Videos, Example
Linear Differential Equations
IVP: First Order Linear Differential Equation: Example 2
Diff Eq: Solving Non-Exact differential equations: Example 2/5
IVP: First Order Linear Differential Equation: Example 1
Linear Differential Equations. Introduction and Example 1.

VIDEO

A01 Introduction to linear systems
Differential Equations: Linear
DE 41 Special Second Order Differential Equations
Solve the differential equation
DE 42 Special Second Order Differential Equations _ Example #5
DIFFERENTIAL EQUATION

COMMENTS

17.1: Second-Order Linear Equations
Consider the differential equation x ″ + 7x ′ + 12x = 0. Both e − 3t and 2e − 3t are solutions (you can check this). However, x(t) = c1e − 3t + c2(2e − 3t) is not the general solution. This expression does not account for all solutions to the differential equation.
Differential Equations (Practice Problems)
We work a couple of examples of solving differential equations involving Dirac Delta functions and unlike problems with Heaviside functions our only real option for this kind of differential equation is to use Laplace transforms. We also give a nice relationship between Heaviside and Dirac Delta functions.
Linear Differential Equation (Solution & Solved Examples)
Examples of linear differential equations are: xdy/dx+2y = x 2 dx/dy - x/y = 2y dy/dx + ycot x = 2x 2 How to solve the first order differential equation? First write the equation in the form of dy/dx+Py = Q, where P and Q are constants of x only Find integrating factor, IF = e ∫Pdx Now write the solution in the form of y (I.F) = ∫Q × I.F C
Differential Equations
The solution to a linear first order differential equation is then y(t) = ∫ μ(t)g(t) dt +c μ(t) (9) (9) y ( t) = ∫ μ ( t) g ( t) d t + c μ ( t) where, μ(t) = e∫p(t)dt (10) (10) μ ( t) = e ∫ p ( t) d t Now, the reality is that (9) (9) is not as useful as it may seem.
Linear Differential Equation
Examples on Linear Differential Equation Example 1: Find the general solution of the differential equation xdy - (y + 2x 2 ).dx = 0 Solution: The give differential equation is xdy - (y + 2x 2 ).dx = 0. This can be simplified to represent the following linear differential equation. dy/dx - y/x = 2x
Linear vs nonlinear differential equation
$\begingroup$ If the ODE has the unknown function and/or its derivative(s) as an argument of a trigonometric, hyperbolic trigonometric, exponential, logarithmic, and/or n-th root function, the ODE is non-linear. If the ODE has a product of the unknown function times any of its derivatives, the ODE is non-linear. If the ODE has the unknown function and/or its derivative(s) with power greater ...
Linear differential equation problems and answers
Linear differential equation problems and answers - We'll provide some tips to help you select the best Linear differential equation problems and answers for ... Examples on Linear Differential Equation Example 1: Find the general solution of the differential equation xdy -(y + 2x2).dx = 0. Solution: The give. Average satisfaction rating 4.7/5 ...
Linear Differential Equation
Linear Differential Equation Examples Example 1: Solve the equation: x \frac {d y} {d x}-2 y=x^ {3} \cos 4 x xdxdy − 2y = x3 cos4x. Answer: Convert the given equation into the standard form (dy / dx) + Py = Q of the linear differential equation. \frac {d y} {d x}-\frac {2} {x} y=x^ {2} \cos 4 x dxdy − x2y = x2 cos4x
Linear differential equations of first order: Solved Example Problems
Maths: Differential Equations: Linear differential equations of first order : Solved Example Problems with Answer, Solution, Formula Example Example 4.24 A firm has found that the cost C of producing x tons of certain product by the equation x dC/dx = 3/x − C and C = 2 when x = 1. Find the relationship between C and x. Solution: Prev Page Next Page
Worked example: linear solution to differential equation
With numbers, you could think of this equation as ax + 5 = 5, where you control the variable a, but the variable x is outside your control, and can be any number whatsoever. The only way you can make this equation true is by making the variable a (the one you control), equal to 0, that way the variable that is outside your control stops messing the equation.
Differential Equations
Learn differential equations for free—differential equations, separable equations, exact equations, integrating factors, and homogeneous equations, and more. ... Second order linear equations Complex and repeated roots of characteristic equation: Second order linear equations Method of undetermined coefficients: ...
PDF LINEAR DIFFERENTIAL EQUATIONS
Therefore, the solution to the initial-value problem is EXAMPLE 3 Solve . SOLUTION The given equation is in the standard form for a linear equation. Multiplying by the integrating factor ... 4 LINEAR DIFFERENTIAL EQUATIONS EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is and the
Stiff equation
In mathematics, a stiff equation is a differential equation for which certain numerical methods for solving the equation are numerically unstable, unless the step size is taken to be extremely small.It has proven difficult to formulate a precise definition of stiffness, but the main idea is that the equation includes some terms that can lead to rapid variation in the solution.
Homogeneous Differential Equations: Solved Example Problems
Example 4.18. If the marginal cost of producing x shoes is given by (3xy + y2 ) dx + (x 2 + xy) dy = 0 and the total cost of producing a pair of shoes is given by ₹12. Then find the total cost function.
Initial Value Problem Differential Equations
In differential equations, initial value problem is often abbreviated IVP. An IVP is a differential equation together with a place for a solution to start, called the initial value. IVPs are often written y ′ = f ( x, y) y ( a) = b where ( a, b) is the point the solution y ( x) must go through.
Linear differential equation
A linear differential equation or a system of linear equations such that the associated homogeneous equations have constant coefficients may be solved by quadrature, which means that the solutions may be expressed in terms of integrals. This is also true for a linear equation of order one, with non-constant coefficients.
Calculus
A diﬀerential equation. dy/dx = f1 (x, y) / f2 (x, y) where f1 and f2 are homogeneous functions of x and y of the same degree, is called a homogeneous diﬀerential equation. Such equations can be solved by taking a new dependent variable v connected with the old one y by the equation y = vx.
Examples of solving linear ordinary differential equations using an
Next: Online quiz: Scalar linear equation problems; Similar pages. Solving linear ordinary differential equations using an integrating factor; An introduction to ordinary differential equations; Ordinary differential equation examples; Exponential growth and decay: a differential equation; Another differential equation: projectile motion
Examples of a nonlinear differential equation
What is an nonlinear ordinary differential equation and give an. An equation in which the maximum degree of a term is 2 or more than two is called a nonlinear equation. + 2x + 1 = 0, 3x + 4y = 5, this is the example of. order now.
Nonhomogeneous differential equation examples
Nonhomogeneous differential equation examples - We'll provide some tips to help you select the best Nonhomogeneous differential equation examples for your ... Explain mathematic problems. Mathematics is the study of numbers, shapes, and patterns. It is used to solve problems. ... NonHomogeneous Second Order Linear Equations (Section 17.2 ...
Differential equation of a polynomial function
Order and Degree of Differential Equations with Examples. Linear differential equations. The general linear ODE of order n is. (1) think of the formal polynomial p(D) as operating on a function y(x), converting. ... Math is a way of solving problems by using numbers and equations.
Ordinary differential equations solving methods
Ordinary differential equation examples. by GE Sjoden Cited by 6 - An Ordinary Differential Equation (ODE) is an equation that defines a relationship between an independent variable x and a dependent variable y ... Inspection Method Variable Separable Method Homogenous Differential Equations Linear Differential Equation. ... To determine what ...
Solving system of differential equations with initial conditions
Step. Calculator Ordinary Differential Equations (ODE) and Systems of ODEs. Calculator applies methods to Without or with initial conditions (Cauchy problem). Do math question. Doing homework can help you learn and understand the material covered in class. Explain mathematic tasks.
Non homogeneous partial differential equations examples
Non homogeneous partial differential equations examples - with a corresponding nonhomogeneous partial differential equation, For example, we might take our ... Nonhomogeneous PDE Problems. The methods for finding the Particular Integrals are the same as those for homogeneous linear equations. If f (D,D') is not homogeneous, then