Solve y"+4y=f(t) text{ subject to } y(0)=0 , y'(0)=2 text{ and } f(t)=begin{cases}0 & t< pi1 & tgeqpiend{cases}

Step 1 Given differential equation $y"+4y=f(t)$ subject to $y(0)=0,y^{\prime }(0)=2$ and $f(t)=\{\begin{array}{ll}0& t<\pi \\ 1& t\ge \pi \end{array}$
$Lf(t)={\int }_0^{\mathrm{\infty }}f(t)e^{-st}dt$
$={\int }_0^{\pi }(0)e^{-st}dt+{\int }_{\pi }^{\mathrm{\infty }}(1)e^{-st}dt$
$=0+{\int }_{\pi }^{\mathrm{\infty }}(1)e^{-st}dt$
$=|\frac{e^{-st}}{-s}{|}_{\pi }^{\mathrm{\infty }}$
$=|\frac{e^{-s(\mathrm{\infty })}}{-s}-\frac{e^{-s(\pi )}}{-s}|$
$=\frac{e^{-\pi t}}{s}$
Step 2
Now taking Laplace transform on both sides of the given differential equation
$L\left\{y^{″}\right\}+4L\left\{y\right\}=L\{f(t)\}$
$s^{2}Y(s)-s.y(0)-y^{\prime }(0)+4Y(s)=\frac{e^{-\pi s}}{s}$
Applying the conditions
$s^{2}Y(s)-s.0-2+4Y(s)=\frac{e^{-\pi s}}{s}$
$Y(s)(s^2+4)=\frac{e^{-\pi s}}{s}+2$
$Y(s)=\frac{e^{-\pi s}}{s(s^2+4)}+\frac{2}{s^2+4}$
Now taking Inverse Laplace transform Step 3
$L^{-1}\{Y(s)\}=L^{-1}\frac{e^{-\pi s}}{s(s^2+4)}+\frac{2}{s^2+4}$
$=L^{-1}\left\{e^{-\pi s}\frac{1}{s(s^2+4)}\right\}+2L^{-1}\left\{\frac{1}{s^2+4}\right\}$
By partial fraction
$\frac{1}{s(s^2+4)}=\frac{1}{4s}-\frac{s}{4(s^2+4)}$
$L^{-1}\{Y(s)\}=L^{-1}\left\{e^{-\pi s}(\frac{1}{4s}-\frac{s}{4(s^2+4)})\right\}+2L^{-1}\left\{\frac{1}{s^2+2^2}\right\}$
Step 4
Apply the rule then $L^{-1}\{e^{-as}F(s)\}=u(t-a)f(t-a)$ and from laplace transform table
$L^{-1}\{\frac{1}{4s}-\frac{s}{4(s^2+4)}\}=\frac{1}{4}-\frac{1}{4}\cos (2t)$
$y(t)=u(1-\pi )(\frac{1}{4}-\frac{1}{4}\cos (2(t-\pi )))+2\left(\frac{1}{2}\right)\sin (2t)$

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