Question

find the response of the function at * (t=4) using Laplace transform (y() + 2y" + y = sint) y(0)=1, y (0)=-2, y"(0)=3 , y"(0)=0

Answer 1

Considering you have four initial conditions (the last of which should probably read
`y'''(0)=0`), I'm assuming the ODE is

`y^((4))(t)+2y''(t)+y(t)=\sin t`

with
`y(0)=1`,
`y'(0)=-2`,
`y''(0)=3`, and
`y'''(0)=0`.

Take the Laplace transform of both sides, denoting the transform of
`y(t)` by
`Y(s)`:

`(s^4Y(s)-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0))+2(s^2Y(s)-sy(0)-y'(0))+Y(s)=\frac1{s^2+1}`

Solve for
`Y(s)`:

`(s^4+2s^2+1)Y(s)-s^3+2s^2-5s+4=\frac1{s^2+1}`

`Y(s)=(1+(s^3-2s^2+5s-4)(s^2+1))/((s^2+1)(s^4+2s^2+1))`

Notice that

`s^4+2s^2+1=(s^2+1)^2`

`\implies Y(s)=(1+(s^3-2s^2+5-4)(s^2+1))/((s^2+1)^3)`

and simplify a bit to get

`Y(s)=(s^5-2s^4+6s^3-6s^2+5s-3)/((s^2+1)^3)`

Decompose
`Y(s)` into partial fractions:

`(s^5-2s^4+6s^3-6s^2+5s-3)/((s^2+1)^3)=(a_0+a_1s)/(s^2+1)+(b_0+b_1s)/((s^2+1)^2)+(c_0+c_1s)/((s^2+1)^3)`

`s^5-2s^4+6s^3-6s^2+5s-3=(a_0+a_1s)(s^2+1)^2+(b_0+b_1s)(s^2+1)+(c_0+c_1s)`

`s^5-2s^4+6s^3-6s^2+5s-3=a_1s^5+a_0s^4+(2a_1+b_1)s^3+(2a_0+b_0)s^2+(a_1+b_1+c_1)s+(a_0+b_0+c_0)`

`\implies\begin{cases}a_1=1\\a_0=-2\\2a_1+b_1=6\\2a_0+b_0=-6\\a_1+b_1+c_1=5\\a_0+b_0+c_0=-3\end{cases}`

`\implies a_0=-2,a_1=1,b_0=-2,b_1=4,c_0=1,c_1=0`

So we have

`Y(s)=(s-2)/(s^2+1)+(4s-2)/((s^2+1)^2)+\frac1{(s^2+1)^3}`

Split up the first term to get two easy inverse transforms:

`L^(-1)\left[\frac s{s^2+1}\right]=\cos t`

`L^(-1)\left[-\frac2{s^2+1}\right]=-2\sin t`

Also split up the second term, but use the convolution theorem, which says

`L\left[(\alpha \ast \beta)(t)\right]=A(s)\cdot B(s)`

where
`A(s)` and
`B(s)` are the Laplace transforms of
`\alpha(t)` and
`\beta(t)`, respectively, and the convolution is defined by

`(\alpha \ast \beta)(t)=\displaystyle\int_0^t\alpha(\tau)\beta(t-\tau)\,\mathrm d\tau`

Take

`A(s)=(4s)/(s^2+1)\text{ and }B(s)=\frac1{s^2+1}`

so that

`\alpha(t)=4\cos t\text{ and }\beta(t)=\sin t`

and their convolution is

`L^(-1)\left[(4s)/((s^2+1)^2)\right]=(\alpha \ast \beta)(t)=2t\sin t`

Next, take

`A(s)=-\frac2{s^2+1}\text{ and }B(s)=\frac1{s^2+1}`

`\implies \alpha(t)=-2\sin t\text{ and }\beta(t)=\sin t`

`\implies L^(-1)\left[-\frac2{(s^2+1)^2}\right]=t\cos t-\sin t`

You can treat the third term similarly, but with an extra step. First compute

`L^(-1)\left[\frac1{(s^2+1)^2}\right]`

by taking

`A(s)=B(s)=\frac1{s^2+1}`

`\implies \alpha(t)=\beta(t)=\sin t`

Then

`L^(-1)\left[\frac1{(s^2+1)^2}\right]=\frac{\sin t-t\cos t}2`

Next, take

`A(s)=\frac1{(s^2+1)^2}\text{ and }B(s)=\frac1{s^2+1}`

`\implies \alpha(t)=\frac{\sin t-t\cos t}2\text{ and }\beta(t)=\sin t`

`\implies L^(-1)\left[\frac1{(s^2+1)^3}\right]=\frac{(3-t^2)\sin t-3t\cos t}8`

Thus we end up with the solution,

`y(t)=(\cos t-2\sin t)+(2t\sin t+t\cos t-\sin t)+\frac{(3-t^2)\sin t-3t\cos t}8`

`\boxed{y(t)=\frac{(8+5t)\cos t+(-21+16t-t^2)\sin t}8}`