Use the Laplace transform to solve the given initial-value problem

$y\prime \prime +2y\prime +y=\delta (t-4)$

$y(0)=0$

$y\prime (0)=0$

Ayaana Buck
2020-11-14
Answered

Use the Laplace transform to solve the given initial-value problem

$y\prime \prime +2y\prime +y=\delta (t-4)$

$y(0)=0$

$y\prime (0)=0$

You can still ask an expert for help

broliY

Answered 2020-11-15
Author has **97** answers

Step 1

Given a differntial equation,

$y\prime \prime +2y\prime +y=\delta (t-4)$

with condition

$y(0)=0$

$y\prime (0)=0$

Step 2

Appling laplace transform on both sides,

$L\{y\prime \prime +2y\prime +y\}=L\{\delta (t-4)\}$

$L\left\{y"\right\}+2L\left\{{y}^{\prime}\right\}+L\left\{y\right\}=L\{\delta (t-4)\}$

Since,

$L\left\{y"\right\}={s}^{2}Y(s)-sy(0)-{y}^{\prime}(0)$

$L\left\{y"\right\}={s}^{2}Y(s)-s(0)-(0)={s}^{2}Y(s)$ and$L\left\{{y}^{\prime}\right\}=sY(s)-y(0)$

$L\left\{{y}^{\prime}\right\}=sY(s)-(0)=sY(s)$

Step 3

Substituting the values,

${s}^{2}Y(s)+2sY(s)+Y(s)={e}^{-4s}$

$({s}^{2}+2s+1)Y(s)={e}^{-4s}$ Thus ,$Y(s)=\frac{{e}^{-4s}}{{s}^{2}+2s+1}$

Step 4

Now, applying the inverse Laplace transform,

${L}^{-1}\{Y(s)\}={L}^{-1}\left\{\frac{{e}^{-4s}}{{s}^{2}+2s+1}\right\}$

$y(t)={L}^{-1}\left\{\frac{{e}^{-4s}}{(s+1{)}^{2}}\right\}$ Since , ${L}^{-1}\left\{\frac{1}{(s+1{)}^{2}}\right\}=t{e}^{-t}$

Now, if inverse laplace transform of F(s) is f(t) then

${L}^{-1}\{{e}^{-at}F(s)\}=H(t-a)f(t-a)$

Thus,

$y(t)=H(t-4)(t-4){e}^{-(t-4)}$

Where H(t-4) is a Heaviside step function.

Step5

Thus, the solution of the given initial value problem is

$y(t)=H(t-4){e}^{-(t-4)}(t-4)$

Given a differntial equation,

with condition

Step 2

Appling laplace transform on both sides,

Since,

Step 3

Substituting the values,

Step 4

Now, applying the inverse Laplace transform,

Now, if inverse laplace transform of F(s) is f(t) then

Thus,

Where H(t-4) is a Heaviside step function.

Step5

Thus, the solution of the given initial value problem is

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