Question

Solve the following I.V.P. by method of Laplace transform:

(i) 𝑦 ′′ + 𝑦 = 𝑓(𝑡), 𝑦(0) = 0, 𝑦 ′ (0) = 0 𝑓(𝑡) = { 2 0 ≤ 𝑡 ≤ 3 3𝑡 − 7 3 < 𝑡 < ∞

(ii) 𝑦 ′′ + 𝑦 = 𝑓(𝑡), 𝑦(0) = 0, 𝑦 ′ (0) = 0 𝑓(𝑡) = { 𝑡^2 0 ≤ 𝑡 ≤ 1 0 1 < 𝑡 < ∞

Answer 1

(i)

$y''+y=f(t),y(0)=0,y'(0)=0$

$f(t)=\begin{cases} 2 &\text{} 0\le t\le 3 \\ 3t-7 &\text{} 3<t<\infty \end{cases}$

Here f(t) is piecewise continuous on$(0,\infty)$, then the derivative theorem exists.

So apply both sides Laplace transformation

$L(y''+y)=L(f(t))$

$L(y''(t))+L(y(t))=L(f(t))$

$s^{2}Y(s)-sY(0)-y'(0)+Y(s)=\int_0^{\infty}{e^{-st}f(t)dt}$

$s^{2}Y(s)-s*0-0+Y(s)=\int_0^{\infty}{e^{-st}f(t)dt}$

$Y(s)(s^{2}+1)=\int_0^{\infty}{e^{-st}f(t)dt}$

$=\int_0^{3}{e^{-st}2dt}+\int_3^{\infty}{e^{-st}(3t-7)dt}$

$=2[\frac{e^{-st}}{-s}]_0^{3}+\int_3^{\infty}{3e^{-st}tdt}-7\int_3^{\infty}{e^{-st}dt}$

$=\frac{-2}{s}[e^{-3s}-1]+3\int_3^{\infty}{te^{-st}dt}-7\int_3^{\infty}{e^{-st}dt}$

$=\frac{-2}{s}[e^{-3s}-1]+3[\frac{te^{-st}}{-s}-\frac{1}{s^{2}}e^{-st}]_3^{\infty}+\frac{7}{s}[e^{-st}]_3^{\infty}$

$=\frac{-2}{s}e^{-3s}+\frac{2}{s}+3[0-0+\frac{e^{-3s}}{s}+\frac{1}{s^{2}}e^{-3s}]+\frac{7}{s}[0-e^{-3s}]$

$=\frac{-2}{s}e^{-3s}+\frac{2}{s}+\frac{3e^{-3s}}{s}+\frac{3e^{-3s}}{s^{2}}+\frac{7e^{-3s}}{s}$

$=e^{-3s}[\frac{-2}{s}+\frac{3}{s}+\frac{3}{s^{2}}-\frac{7}{s}]+\frac{2}{s}$

$=e^{-3s}[\frac{-2s+3s+3-7s}{s^{2}}]+\frac{2}{s}$

$=e^{-3s}[\frac{3-6s}{s^{2}}]+\frac{2}{s}$

$Y(s)=\frac{1}{s^{2}+1}[(e^{-3s}(\frac{3-6s}{s^{2}}))+\frac{2}{s}]$

Where Y(s) is the Laplace transformation of y(t)

(ii)

$y''+y=f(t),y(0)=0,y'(0)=0$

$f(t)=\begin{cases} t^{2} &\text{} 0\le t\le 1 \\ 0 &\text{} 1<t<\infty \end{cases}$

So we apply Laplace transformation on both sides

$L(y''+y)=L(f(t))$

$L(y''(t))+L(y(t))=L(f(t))$

$s^{2}Y(s)-sY(0)-y'(0)+Y(s)=\int_0^{\infty}{e^{-st}f(t)dt}$

$s^{2}Y(s)-0-0+Y(s)=\int_0^{\infty}{e^{-st}f(t)dt}$

Where f(t) is a piecewise continuous function

$Y(s)(s^{2}+1)=\int_0^{1}{t^{2}e^{-st}dt}+\int_1^{\infty}{0*e^{-st}dt}$

$=\int_0^{1}{t^{2}e^{-st}dt}+0$

$=[t^{2}\frac{e^{-st}}{-s}]_0^{1}-\int_0^{1}{2t\frac{e^{-st}}{-s}dt}$

$=[1\frac{e^{-s}}{-s}+0]-\frac{2}{s}[t\frac{e^{-st}}{-s}-\frac{1}{s^{2}}e^{-st}]_0^{1}$

$=-\frac{e^{-s}}{s}-\frac{2}{s}[\frac{e^{-s}}{-s}-\frac{e^{-s}}{s^{2}}+\frac{1}{s^{2}}]$

$=-\frac{e^{-s}}{s}+\frac{2e^{-s}}{s^{2}}+\frac{2e^{-s}}{s^{3}}-\frac{2}{s^{3}}$

$=\frac{e^{-s}}{s}[-1+\frac{2}{s}+\frac{2}{s^{2}}]-\frac{2}{s^{3}}$

$Y(s)=\frac{e^{-s}}{(s^{2}+1)s}[-1+\frac{2}{s}+\frac{2}{s^{2}}]-\frac{2}{s^{3}(s^{2}+1)}$

Where Y(s) is the Laplace transformation of y(t).