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17Calculus - Verify Differential Equations Solutions

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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.

You Can Have an Amazing Memory: Learn Life-Changing Techniques and Tips from the Memory Maestro

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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Really UNDERSTAND Differential Equations

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