\begin{equation*}
y^\prime = 3y \quad \text{and} \quad y^\prime  3y = 0\text{.}
\end{equation*}
Meaning that you are free to use either form when verifying solutions. For beginners, it is often easier to verify solutions using the second form, as you just have to show the leftside of equation simplifies to zero, giving you \(0=0\text{.}\)Section 2.2 Verifying Solutions
In the previous section, we discussed what it means for a function to be a solution to a differential equation. Now, we turn our attention to how we can verify that a proposed function is indeed a solution.
The process of verifying a solution is straightforward: you substitute the proposed function into the differential equation and check whether the equation holds true. This simple test helps ensure that the function satisfies the relationship defined by the differential equation.
Remember, the following differential equations are identical: Let’s look at a few examples.
Note 17. Tip: Before verifying, move all terms to one side.
Example 18.
\(\ \ \)Verify the given solutions for each differential equation.A. Verify that \(\ y= 2x^2\ \) is a solution to \(\ xy'  2x^2 = y\text{.}\)
Solution 1.
Let’s begin by moving all terms to the leftside of the equation.
\begin{equation*}
xy'  2x^2  y = 0
\end{equation*}
To reduce errors and break the problem into more manageable steps, we will compute and simplify the derivatives appearing in the equation.
\begin{equation*}
y = 2x^2, \quad y' = 4x
\end{equation*}
Finally, we substitute in \(y\) and \(y'\) and simplify.
\begin{align*}
xy'  2x^2  y =\amp\ 0 \\
x\left(4x\right)  2x^2  \left( 2x^2 \right) =\amp\ 0 \\
4x^2  2x^2  2x^2 =\amp\ 0 \\
0 =\amp\ 0 \quad \leftarrow \text{true}
\end{align*}
Therefore, \(y= 2x^2\) is a solution to \(xy'  2x^2 = y.\)
B. Verify that \(\ P = \sin t\ \) is a solution to \(\ 2P'' + P = \sin t\text{.}\)
Solution 2.
As before, we compute \(P''\) since it appears in the equation and move all terms to the leftside.
\begin{equation*}
P = \sin t, \quad P' = \cos t, \quad P'' = \sin t
\end{equation*}
Plug \(P\) and \(P''\) in and simplify.
\begin{align*}
2P'' + P  \sin t =\amp\ 0 \\
2\left( \sin t \right) + \sin t  \sin t =\amp\ 0 \\
2\sin t + \cancel{\sin t}  \cancel{\sin t} =\amp\ 0 \\
2\sin t =\amp\ 0\quad \leftarrow \text{false}
\end{align*}
Therefore, \(P = \sin t\) is not a solution to \(2P'' + P = \sin t.\)
Example 19.
\(\ \ \)Verify that both \(y= 2e^{3t}\) and \(y= e^{4t}\) are solutions to
\begin{equation*}
\quad y''  y'  12y = 0\text{.}
\end{equation*}
Solution 1. \(\quad y= 2e^{3t}\)
Find \(y'\) and \(y''\) since they appear in the equation.
\begin{align*}
y =\amp\ 2e^{3t} \\
y' =\amp\ 2e^{3t}\left(3\right) = 6e^{3t} \\
y'' =\amp\ 6e^{3t}\left(3\right) = 18e^{3t}
\end{align*}
Plug \(y, y',\) and \(y''\) into the DE and simplify.
\begin{align*}
y''  y'  12y =\amp\ 0 \\
18e^{3t}  \left( 6e^{3t} \right)  12\left( 2e^{3t} \right) =\amp\ 0 \\
18e^{3t} + 6e^{3t}  24e^{3t} =\amp\ 0 \\
0 =\amp\ 0 \quad \leftarrow \text{true}
\end{align*}
Therefore, \(y= 2e^{3t}\) is a solution to \(y''  y'  12y = 0.\)
Solution 2. \(\quad y= e^{4t}\)
Find \(y'\) and \(y''\) since they appear in the DE.
\begin{align*}
y =\amp\ e^{4t} \\
y' =\amp\ e^{4t}\left(4\right) = 4e^{4t} \\
y'' =\amp\ 4e^{4t}\left(4\right) = 16e^{4t}
\end{align*}
Plug \(y, y',\) and \(y''\) into the DE and simplify.
\begin{align*}
y''  y'  12y =\amp\ 0 \\
16e^{4t}  4e^{4t}  12\left( e^{4t} \right) =\amp\ 0 \\
16e^{4t}  4e^{4t}  12e^{4t} =\amp\ 0 \\
0 =\amp\ 0 \quad \leftarrow \text{true}
\end{align*}
Therefore, \(y= e^{4t}\) is a solution to \(y''  y'  12y = 0.\)
Example 20.
\(\ \ \)Verify that \(\ds\ y = 3\sin(x^2) \) is a solution to
\begin{equation*}
y'  xy'' = 12x^2\sin(x^2)\text{.}
\end{equation*}
Solution.
Find \(y'\) and \(y''\) since they appear in the equation and move all terms to the leftside.
\begin{align*}
y =\amp\ 3\sin(x^2) \\
y' =\amp\ 3\cos(x^2) \cdot 2x = 6x\cos(x^2) \\
y'' =\amp\ 6\cos(x^2)  12x\sin(x^2) = 6\cos(x^2)  12x\sin(x^2)
\end{align*}
Plug in \(y, y',\) and \(y''\) and simplify.
\begin{align*}
y'  xy''  12x^2\sin(x^2)
=\amp\ 0 \\
6x\cos(x^2)  x\left( 6\cos(x^2)  12x\sin(x^2) \right)  12x^2\sin(x^2)
=\amp\ 0 \\
\cancel{6x\cos(x^2)}  \cancel{6x\cos(x^2)}  \cancel{12x^2\sin(x^2)}  \cancel{12x^2\sin(x^2)}
=\amp\ 0\\
0 =\amp\ 0
\end{align*}
Therefore, \(y= 3\sin(x^2)\) is a solution to \(y'  12x^2\sin(x^2) = xy''.\)
You can even verify that a function is a solution to a differential equation when the function contains constants as the following example shows.
Example 21.
\(\ \ \)Verify that \(\ds y = \frac{1}{2}x^2 + c_1 x + c_2\) is a solution to \(\ y'' = 1\text{.}\)Solution.
Move terms to left: \(\quad y''  1 = 0\)
Find \(y''\) since it appears in the DE.
\begin{align*}
y =\amp\ \frac{1}{2}x^2 + c_1 x + c_2 \\
y' =\amp\ x + c_1 \\
y'' =\amp\ 1
\end{align*}
Plug \(y''\) into the DE and simplify.
\begin{align*}
y''  1 =\amp\ 0 \\
1  1 =\amp\ 0 \\
0 =\amp\ 0
\end{align*}
Therefore, \(\ds y = \frac{1}{2}x^2 + c_1 x + c_2\) is a solution to \(y'' = 1\text{.}\)
^{ 2 }
In fact, it is the general solution, but we don’t show that here.
As you work through more complex differential equations, this verification process becomes a valuable tool. In the next section, we’ll discuss the different types of solutions you will encounter and how you can visualize them.
Reading Questions Check your Understanding
1. Moving all terms of a differential equation to one side of the equation is a required step for verifying the solution to a differential equation..

Moving all terms of a differential equation to one side of the equation is a required step for verifying the solution to a differential equation.
 True
 Incorrect. Read the note provided in this section.
 False
 Correct! This is not a required step, but it can sometimes simplify the process.
2. \(y = x^2 + 3\) is a solution to \(\ds\frac{dy}{dx}  3 = 2x\).

\(y = x^2 + 3\) is a solution to the differential equation \(\displaystyle \frac{dy}{dx}  3 = 2x\text{.}\)
 True
 Incorrect. \(y = x^2 + 3\) is not a solution since\begin{align*} \frac{dy}{dx}  3 =\amp\ 2x \\ \frac{d}{dx}\left[x^2 + 3\right]  3 =\amp\ 2x \\ 2x  3 =\amp\ 2x \quad \leftarrow \text{false} \end{align*}
 False
 Correct! \(y = x^2 + 3\) is not a solution since\begin{align*} \frac{dy}{dx}  3 =\amp\ 2x \\ \frac{d}{dx}\left[x^2 + 3\right]  3 =\amp\ 2x \\ 2x  3 =\amp\ 2x \quad \leftarrow \text{false} \end{align*}
3. List the steps needed to verify \(s = \cos(3x)\) is a solution to \(s''  6s' + 9s = 0\).
4. A differential equation has one solution.
 True
 Incorrect. Example 19 shows a differential equation with two solutions.
 False
 Correct!
5. Consider the differential equation with missing righthand side:
\begin{equation*}
y''  \frac{4}{x}y' = \fillinmath{XXXXX}\text{.}
\end{equation*}
Assuming \(y = 2x^3\) is a solution to this equation, which of the following is a possible righthand side?
\begin{equation*}
y''  \frac{4}{x}y' = \fillinmath{XXXXX}\text{.}
\end{equation*}
Assuming \(y = 2x^3\) is a solution to this equation, which of the following is a possible righthand side?
 \(2x^3 + 4x^2 \)
 Incorrect. Plug \(y'\) and \(y''\) into the leftside and simplify.
 \(0 \)
 Incorrect. Plug \(y'\) and \(y''\) into the leftside and simplify.
 \(12x \)
 Correct! Plugging \(y' = 6x^2\) and \(y'' = 12x\) into the leftside gives\begin{align*} (12x)  \frac{4}{x}(6x^2) =\amp\ \fillinmath{XXXXX} \\ 12x  24x =\amp\ \fillinmath{XXXXX} \\ 12x =\amp\ \fillinmath{XXXXX} \quad \leftarrow \text{true} \end{align*}and to get a true statement, the righthand side must be \(12x\text{.}\)
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