Roger Smith
2021-12-18
Answered

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

Answered 2021-12-19
Author has **40** answers

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,$\frac{dy}{dx}=\frac{d}{dx}\left({e}^{-3x}\right)={e}^{-3x}(-3)=-3{e}^{-3x}$

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{d}{dx}-3\left({e}^{-3x}\right)=(-3)\frac{d}{dx}\left({e}^{-3x}\right)=-3\left({e}^{-3x}(-3)\right)=9{e}^{-3x}$

$-6y=-6{e}^{-3x}$

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.

To verify the function

That is the function

Step 2

Now,

Therefore,

Hence, equation (1) satisfies.

Therefore, the given function

Orlando Paz

Answered 2021-12-20
Author has **42** answers

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 $\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}-6y=0$

On differentiating with x, we get

On differentiating again with x, we get

Now let's see what is the value of

Conclusion: Therefore,

RizerMix

Answered 2021-12-29
Author has **438** answers

Differentiating with x

Again differentiating,

Substituting in

Hence

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