Finding the Laplace transform of the solution of the given IVP

Question

Find the the Laplace transform $Y(s)$ of the solution of the given initial value problem $$y''+y=\begin{cases}t & 0 < t < 1 \\ 0 & 1 < t < \infty \end{cases}$$

$$y(0)=0$$

$$y'(0)=0$$

What i tried

The laplace transform becomes $$Y(S)+sy(0)+y'(0)=\begin{cases}\frac{1}{s^2} & 0 < \frac{1}{s^2} < 1 \\ 0 & 1 < \frac{1}{s^2} < \infty \end{cases}$$ Since $$y(0)=0$$

$$y'(0)=0$$ the equation becomes $$Y(S)=\begin{cases}\frac{1}{s^2} & 0 < \frac{1}{s^2} < 1 \\ 0 & 1 < \frac{1}{s^2} < \infty \end{cases}$$ AM i correct. COuld anyone please explain. Thanks

Answer 1

The Laplace Transform of the derivative of a function $y(t)$ with initial value $y(o)$ is $sY(s)-y(0)$ while the Laplace Transform of the second derivative of a function $y(t)$ with initial values $y(o)$ and $y'(0)$ is $s^2Y(s)-sy(0)-y'(0)$.

So, taking the Laplace Transform of the RHS of the ODE yields $$(s^2+1)Y(s)-sy(o)-y'(0)$$ For the RHS, the Laplace Transform is given by $$\int_0^{1} te^{-st}dt=-\frac{e^{-s}}{s}-\frac{e^{-s}-1}{s^2}$$ Equating sides reveals that $$(s^2+1)Y(s)-sy(o)-y'(0)=-\frac{e^{-s}}{s}-\frac{e^{-s}-1}{s^2}$$

Answer 2

Applying Laplace transform you will get

$s^2Y(s)-sy(0)-y'(0)=\int_0^1e^{-st}tdt$

after using ICs and evaluating integral,

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

Finally applying inverse Laplace Transform

$y(t)=L^{-1}\{Y(s)\}=t-u_1(t)(t-\cos(1-t)+\sin(1-t))-\sin(t)$

where $u_1(t)=0$ for $t\leq1$ and $u_1(t)=1$ for $t>1$.