Question

Solve the initial value problem using Laplace transforms.

y''-7y'+10y=9 cos(t)+7sin(t)

y(0)=5, y'(0)=-4

Answer 1

Let
`\mathcal L_s\{y(t)\}=Y(s)` denote the Laplace transform of
`y(t)`. Recall that


`\mathcal L_s\{y''(t)\}=s^2Y(s)-sy(0)-y'(0)`

`\mathcal L_s\{y'(t)\}=sY(s)-y(0)`

`\mathcal L_s\{\cos t\}=\frac s{s^2+1}`

`\mathcal L_s\{\sin t\}=\frac1{s^2+1}`

Taking the transform of both sides yields


`\bigg(s^2Y(s)-5s+4\bigg)-7\bigg(sY(s)-5\bigg)+10Y(s)=(9s+7)/(s^2+1)`

and solving for
`Y(s)` gives


`Y(s)=((9s+7)/(s^2+1)+5s-39)/(s^2-7s+10)`

`Y(s)=(5s^3-39s^2+14s-32)/((s^2+1)(s^2-7s+10))`

`Y(s)=(5s^3-39s^2+14s-32)/((s^2+1)(s-5)(s-2))`

`Y(s)=-\frac4{s-5}+\frac8{s-2}+\frac s{s^2+1}`

Take the inverse transform and you're done:


`y(t)=\mathcal L^(-1)_t\{Y(s)\}=-4e^(5t)+8e^(2t)+\cos t`