# Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

## Section2.2Verifying 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.

### Note17.Tip: Before verifying, move all terms to one side.

Remember, the following differential equations are identical:
\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 left-side of equation simplifies to zero, giving you $$0=0\text{.}$$
Let’s look at a few examples.

### Example18.

$$\ \$$Verify the given solutions for each differential equation.
A. Verify that $$\ y= 2x^2\$$ is a solution to $$\ xy' - 2x^2 = y\text{.}$$
Solution 1.
Let’s begin by moving all terms to the left-side 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''$$ since it appears in the equation and move all terms to the left-side.
\begin{equation*} P = \sin t, \quad P' = \cos t, \quad P'' = -\sin t \end{equation*}
Plug $$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.$$

### Example19.

$$\ \$$Verify that both $$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.$$

### Example20.

$$\ \$$Verify that $$\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''$$ since they appear in the equation and move all terms to the left-side.
\begin{align*} y =\amp\ 3\sin(x^2) \\ y' =\amp\ 3\cos(x^2) \cdot 2x = 6x\cos(x^2) \\ y'' =\amp\ 6\cos(x^2) - 12x\sin(x^2) = 6\cos(x^2) - 12x\sin(x^2) \end{align*}
Plug in $$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.

### Example21.

$$\ \$$Verify that $$\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.

#### 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*}

#### 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 right-hand 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 right-hand side?

• $$2x^3 + 4x^2$$
• Incorrect. Plug $$y'$$ and $$y''$$ into the left-side and simplify.
• $$0$$
• Incorrect. Plug $$y'$$ and $$y''$$ into the left-side and simplify.
• $$-12x$$
• Correct! Plugging $$y' = 6x^2$$ and $$y'' = 12x$$ into the left-side 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 right-hand side must be $$-12x\text{.}$$