## 17Calculus - Verify Differential Equations Solutions

##### 17Calculus

This page explains how to verify that we have a solution to a differential equation.

If you would like a full lecture on this, we recommend this video.

### Prof Leonard - Checking Solutions in Differential Equations [30mins-57secs]

video by Prof Leonard

Before we start discussing how to actually solve differential equations, let's talk about how to check that we have a solution. Remember from algebra, that if you have an equation, you can check if a specific value solves the equation. Let's look at a quick example.

Example 1

Show that $$x=1$$ solves the equation $$x^2-1 = 0$$.

To solve this we just plug in $$1$$ for $$x$$ in the equation.

So $$1^2 - 1 = 1-1 = 0 ~ \to ~ 0 = 0$$

What we have done here is plugged in $$x=1$$ on the left side of the equal sign and shown that the result is equal to the right side.

Note that $$x=1$$ is one of the solutions to the equation. There is another solution at $$x=-1$$ but we don't need to worry about that one in this problem. We were asked to just verify $$x=1$$ and that's all we needed to do.

To verify a solution to a differential equation, we do the same thing, i.e. we are given a solution, we plug it into the differential equation and verify that resulting equation is true. Let's look at an example.

Example 2

Verify that $$y=x+1$$ is a solution to $$(x+1)y' - y = 0$$.

The idea here is to plug in $$y=x+1$$ into the left side of differential equation to show that we get zero. Looking at the differential equation, the second term is just $$y$$, so that's easy. But the first term has a $$y'$$. Before we go on, we must find $$y'$$ from $$y=x+1$$. The calculation is $$y=x+1 \to y' = 1 + 0 = 1$$. So we have $$y'=1$$. Plugging $$y=x+1$$ and $$y'=1$$ into the differential equation, we have
$$\begin{array}{rcl} (x+1)y' - y & = & 0 \\ (x+1)(1) - (x+1) & = & 0 \\ x+1 -x-1 & = & 0 \\ 0 & = & 0 \end{array}$$
Since $$0 = 0$$ is always true, we can say that $$y=x+1$$ is a solution to the differential equation $$(x+1)y' - y = 0$$.

You may also be asked if a function satisfies a differential equation. In this case, we follow the same steps but at the end, if what we end up with does not hold, then the function is NOT a solution.

Let's work a few practice problems verifying differential equations solutions.

Practice

Verify that $$y=2e^t/3 + e^{-2t}$$ is a solution to $$y' + 2y = 2e^t$$

Problem Statement

Verify that $$y=2e^t/3 + e^{-2t}$$ is a solution to $$y' + 2y = 2e^t$$

Solution

### blackpenredpen - 3649 video solution

video by blackpenredpen

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Verify that $$(-1/2)t\cos t$$ is a solution to $$\displaystyle{ \frac{d^2y}{dt^2} + y = \sin t }$$

Problem Statement

Verify that $$(-1/2)t\cos t$$ is a solution to $$\displaystyle{ \frac{d^2y}{dt^2} + y = \sin t }$$

Solution

### blackpenredpen - 3650 video solution

video by blackpenredpen

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Verify that $$y = x^4/16$$ is a solution to $$dy/dx = xy^{1/2}$$

Problem Statement

Verify that $$y = x^4/16$$ is a solution to $$dy/dx = xy^{1/2}$$

Solution

### MIP4U - 3651 video solution

video by MIP4U

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Verify that $$y=4\cos(2x) + 6\sin(2x)$$ is a solution to $$y''+4y=0$$

Problem Statement

Verify that $$y=4\cos(2x) + 6\sin(2x)$$ is a solution to $$y''+4y=0$$

Solution

### MIP4U - 3652 video solution

video by MIP4U

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Verify that $$y=e^{-x} + 2xe^{-x}$$ is a solution to $$d^2y/dx^2 + 2dy/dx + y = 0$$

Problem Statement

Verify that $$y=e^{-x} + 2xe^{-x}$$ is a solution to $$d^2y/dx^2 + 2dy/dx + y = 0$$

Solution

### MIP4U - 3653 video solution

video by MIP4U

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Is $$y=3\sin(2x) + e^{-x}$$ a solution to $$y''+4y = 5e^{-x}$$?

Problem Statement

Is $$y=3\sin(2x) + e^{-x}$$ a solution to $$y''+4y = 5e^{-x}$$?

Solution

### blackpenredpen - 3655 video solution

video by blackpenredpen

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Which of the following is a solution to $$dy/dx = x^4$$?
(a) $$y = x^5/5+x$$
(b) $$y=4^x + 5$$
(c) $$x^5/5 + \pi$$

Problem Statement

Which of the following is a solution to $$dy/dx = x^4$$?
(a) $$y = x^5/5+x$$
(b) $$y=4^x + 5$$
(c) $$x^5/5 + \pi$$

Solution

### PatrickJMT - 3654 video solution

video by PatrickJMT

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For the differential equation $$\displaystyle{ \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2 }$$, determine if $$y = -\ln(2-x)$$ and/or $$y = -\sqrt{2-x}$$ are solutions. Note: It is possible that one, both or neither are solutions.

Problem Statement

For the differential equation $$\displaystyle{ \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2 }$$, determine if $$y = -\ln(2-x)$$ and/or $$y = -\sqrt{2-x}$$ are solutions. Note: It is possible that one, both or neither are solutions.

Solution

### blackpenredpen - 3659 video solution

video by blackpenredpen

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For what values of $$r$$ will $$y=e^{rx}$$ satisfy $$2y'' + y' - y = 0$$?

Problem Statement

For what values of $$r$$ will $$y=e^{rx}$$ satisfy $$2y'' + y' - y = 0$$?

Solution

### blackpenredpen - 3660 video solution

video by blackpenredpen

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