Respuesta :
Considering you have four initial conditions (the last of which should probably read [tex]y'''(0)=0[/tex]), I'm assuming the ODE is
[tex]y^{(4)}(t)+2y''(t)+y(t)=\sin t[/tex]
with [tex]y(0)=1[/tex], [tex]y'(0)=-2[/tex], [tex]y''(0)=3[/tex], and [tex]y'''(0)=0[/tex].
Take the Laplace transform of both sides, denoting the transform of [tex]y(t)[/tex] by [tex]Y(s)[/tex]:
[tex](s^4Y(s)-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0))+2(s^2Y(s)-sy(0)-y'(0))+Y(s)=\dfrac1{s^2+1}[/tex]
Solve for [tex]Y(s)[/tex]:
[tex](s^4+2s^2+1)Y(s)-s^3+2s^2-5s+4=\dfrac1{s^2+1}[/tex]
[tex]Y(s)=\dfrac{1+(s^3-2s^2+5s-4)(s^2+1)}{(s^2+1)(s^4+2s^2+1)}[/tex]
Notice that
[tex]s^4+2s^2+1=(s^2+1)^2[/tex]
[tex]\implies Y(s)=\dfrac{1+(s^3-2s^2+5-4)(s^2+1)}{(s^2+1)^3}[/tex]
and simplify a bit to get
[tex]Y(s)=\dfrac{s^5-2s^4+6s^3-6s^2+5s-3}{(s^2+1)^3}[/tex]
Decompose [tex]Y(s)[/tex] into partial fractions:
[tex]\dfrac{s^5-2s^4+6s^3-6s^2+5s-3}{(s^2+1)^3}=\dfrac{a_0+a_1s}{s^2+1}+\dfrac{b_0+b_1s}{(s^2+1)^2}+\dfrac{c_0+c_1s}{(s^2+1)^3}[/tex]
[tex]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)[/tex]
[tex]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)[/tex]
[tex]\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}[/tex]
[tex]\implies a_0=-2,a_1=1,b_0=-2,b_1=4,c_0=1,c_1=0[/tex]
So we have
[tex]Y(s)=\dfrac{s-2}{s^2+1}+\dfrac{4s-2}{(s^2+1)^2}+\dfrac1{(s^2+1)^3}[/tex]
Split up the first term to get two easy inverse transforms:
[tex]L^{-1}\left[\dfrac s{s^2+1}\right]=\cos t[/tex]
[tex]L^{-1}\left[-\dfrac2{s^2+1}\right]=-2\sin t[/tex]
Also split up the second term, but use the convolution theorem, which says
[tex]L\left[(\alpha \ast \beta)(t)\right]=A(s)\cdot B(s)[/tex]
where [tex]A(s)[/tex] and [tex]B(s)[/tex] are the Laplace transforms of [tex]\alpha(t)[/tex] and [tex]\beta(t)[/tex], respectively, and the convolution is defined by
[tex](\alpha \ast \beta)(t)=\displaystyle\int_0^t\alpha(\tau)\beta(t-\tau)\,\mathrm d\tau[/tex]
Take
[tex]A(s)=\dfrac{4s}{s^2+1}\text{ and }B(s)=\dfrac1{s^2+1}[/tex]
so that
[tex]\alpha(t)=4\cos t\text{ and }\beta(t)=\sin t[/tex]
and their convolution is
[tex]L^{-1}\left[\dfrac{4s}{(s^2+1)^2}\right]=(\alpha \ast \beta)(t)=2t\sin t[/tex]
Next, take
[tex]A(s)=-\dfrac2{s^2+1}\text{ and }B(s)=\dfrac1{s^2+1}[/tex]
[tex]\implies \alpha(t)=-2\sin t\text{ and }\beta(t)=\sin t[/tex]
[tex]\implies L^{-1}\left[-\dfrac2{(s^2+1)^2}\right]=t\cos t-\sin t[/tex]
You can treat the third term similarly, but with an extra step. First compute
[tex]L^{-1}\left[\dfrac1{(s^2+1)^2}\right][/tex]
by taking
[tex]A(s)=B(s)=\dfrac1{s^2+1}[/tex]
[tex]\implies \alpha(t)=\beta(t)=\sin t[/tex]
Then
[tex]L^{-1}\left[\dfrac1{(s^2+1)^2}\right]=\dfrac{\sin t-t\cos t}2[/tex]
Next, take
[tex]A(s)=\dfrac1{(s^2+1)^2}\text{ and }B(s)=\dfrac1{s^2+1}[/tex]
[tex]\implies \alpha(t)=\dfrac{\sin t-t\cos t}2\text{ and }\beta(t)=\sin t[/tex]
[tex]\implies L^{-1}\left[\dfrac1{(s^2+1)^3}\right]=\dfrac{(3-t^2)\sin t-3t\cos t}8[/tex]
Thus we end up with the solution,
[tex]y(t)=(\cos t-2\sin t)+(2t\sin t+t\cos t-\sin t)+\dfrac{(3-t^2)\sin t-3t\cos t}8[/tex]
[tex]\boxed{y(t)=\dfrac{(8+5t)\cos t+(-21+16t-t^2)\sin t}8}[/tex]
