DE-BOOK Verifying Solutions

Section 3.2 Verifying Solutions

Verifying that a given function is a solution to a differential equation amounts to showing that it satisfies the equation. This typically involves taking any necessary derivatives of the function, substituting them into the equation, and simplifying both sides to determine whether they match. Here are some examples to illustrate this process.

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🌌 Example 23.

Verify that \(y = 2x^2\ \) is a solution to \(\ \ xy' - 2x^2 = y\text{.}\)

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Solution.

First, compute \(y' = 4x\) and plug both \(y = 2x^2\) and \(y' = 4x\) into the equation:

\begin{align*} x\left( 4x \right) - 2x^2 \amp = \left( 2x^2 \right) \\ 4x^2 - 2x^2 \amp = 2x^2 \\ 2x^2 \amp = 2x^2 \quad \end{align*}

This shows that \(y = 2x^2\) satisfies the equation and proves it is a solution.

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🌌 Example 24.

Verify that \(P = \sin t\ \) is a solution to \(\ \ 2P'' + P = \sin t\text{.}\)

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Solution.

Since \(P''\) appears, compute the derivatives first:

\begin{equation*} P = \sin t, \quad P' = \cos t, \quad P'' = -\sin t \end{equation*}

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Now substitute in the dependent variable along with its derivatives:

\begin{align*} 2\left( -\sin t \right) + \left( \sin t \right) \amp = \sin t\\ -2\sin t + \sin t \amp = \sin t \\ -\sin t \amp \ne \sin t \quad \end{align*}

Since the function on the left is different from the function on the right, \(P = \sin t\) does not satisfy the equation and is not a solution.

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As you will soon learn, differential equations have many solutions and all of them share a common form, as the next example shows.

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🌌 Example 25.

Show that the functions

\begin{equation*} y = e^{2x},\quad y = -5e^{2x},\quad y = \pi e^{2x},\quad y = 0 \end{equation*}

are all solutions to the differential equation

\begin{equation*} y' - 2y = 0\text{.} \end{equation*}

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Solution.

Rather than verify each function separately, notice that they are all of the form \(y = c e^{2x}\text{,}\) where \(c\) is a constant. Let’s verify the general case.

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Substitute \(y = c e^{2x}\) into the equation:

\begin{equation*} y' - 2y = (c e^{2x})' - 2(c e^{2x}) = 2c e^{2x} - 2c e^{2x} = 0 \quad \end{equation*}

So \(y = c e^{2x}\) is a solution for any constant \(c\text{.}\) In particular, this includes the functions

\begin{equation*} y = e^{2x},\quad y = -5e^{2x},\quad y = \pi e^{2x},\quad y = 0 \end{equation*}

for the constants \(c = 1, -5, \pi, 0\) respectively.

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As you saw in the previous example, solutions can differ by a constant. Some solutions even involve multiple constants, as the next example shows.

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🌌 Example 26.

Verify that \(\ y = c_1\ x^2 + c_2 - \ln x\ \) is a solution to

\begin{equation*} x^2y'' - xy' = 2\text{.} \end{equation*}

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Solution.

Compute the needed derivatives:

\begin{equation*} y = c_1\ x^2 + c_2 - \ln x \ \ \Rightarrow \ \ y' = 2 c_1\ x - \frac{1}{x} \ \ \Rightarrow \ \ y'' = 2 c_1 + \frac{1}{x^2} \end{equation*}

and substitute into the equation:

\begin{align*} x^2\left( 2 c_1 + \frac{1}{x^2} \right) - x\left( 2 c_1\ x - \frac{1}{x} \right) \amp = 2\\ 2 c_1\ x^2 + 1 - 2 c_1\ x^2 + 1 \amp = 2\\ 2 \amp = 2 \quad \end{align*}

Thus, \(\ y = c_1\ x^2 + c_2 - \ln x\ \) is a solution to \(x^2y'' - xy' = 2\text{.}\)

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