# Verify that the function y=e^{-3x} is a solution to the

Verify that the function $y={e}^{-3x}$ is a solution to the differential equation ${d}^{2}\frac{y}{{dx}^{2}}+\frac{dy}{dx}-6y=0$?
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Becky Harrison
Step 1
To verify the function $y={e}^{-3x}$ is a solution of the differential equations $\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}-6y=0$.
That is the function $y={e}^{-3x}$ satisfies the equation $\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}-6y=0$ ...(1)
Step 2
Now, $\frac{dy}{dx}=\frac{d}{dx}\left({e}^{-3x}\right)={e}^{-3x}\left(-3\right)=-3{e}^{-3x}$
$\frac{{d}^{2}y}{{dx}^{2}}=\frac{d}{dx}-3\left({e}^{-3x}\right)=\left(-3\right)\frac{d}{dx}\left({e}^{-3x}\right)=-3\left({e}^{-3x}\left(-3\right)\right)=9{e}^{-3x}$
$-6y=-6{e}^{-3x}$
Therefore, $\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}-6y$
$=9{e}^{-3x}-3{e}^{-3x}-6{e}^{-3x}$
$=9{e}^{-3x}-9{e}^{-3x}$
$=0$
Hence, equation (1) satisfies.
Therefore, the given function $y={e}^{-3x}$ is a solution to the given differential equation.
###### Not exactly what you’re looking for?
Orlando Paz
Given, $y={e}^{-3x}$
On differentiating with x, we get
$\frac{dy}{dx}=-3{e}^{-3x}$
On differentiating again with x, we get
$\frac{{d}^{2}y}{{dx}^{2}}=9{e}^{-3x}$
Now let's see what is the value of $\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}-6y$
$=9{e}^{-3x}-3{e}^{-3x}-6{e}^{-3x}$
$=0$
Conclusion: Therefore, $y={e}^{-3x}$ is the solution of $\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}-6y=0$
###### Not exactly what you’re looking for?
RizerMix

$y={e}^{-3x}$
Differentiating with x
$\frac{dy}{dx}=-3{e}^{-3x}$
Again differentiating, $\frac{{d}^{2}y}{d{x}^{2}}=9{e}^{-3x}$
Substituting in $\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}-6y$
$9{e}^{-3x}-3{e}^{-3x}-\left(6{e}^{-3x}\right)=0$
Hence $y={e}^{-3x}$ is a solution.