Question

Solve the initial value problem using Laplace transforms. y"+y=f(t) , y(0)=0 , y'(0)=1

Laplace transform
ANSWERED
asked 2021-09-06

Solve the initial value problem using Laplace transforms.
\(y"+y=f(t) , y(0)=0 , y'(0)=1\)
Here
\(f(t)=\begin{cases}0 & 0\leq t<3\pi\\1 & t\geq 3\pi\end{cases}\)

Expert Answers (1)

2021-09-07

Given differential equation
\(y"+y=f(t), y(0)=0 , y'(0)=1\)
Here
\(f(t)=\begin{cases}0 & 0\leq t<3\pi\\1 & t\geq 3\pi\end{cases}\)
To find: Solve the initial value problem using Laplace Transform
We begin with taking Laplace transformation on both sides,
\(\displaystyle{L}{\left({y}\text{}{y}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}\)
\(\displaystyle{L}{\left({y}\text{}+{L}{\left({y}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}\right.}\)
Note: \(\displaystyle{L}{\left({y}\text{}={s}^{{2}}{L}{\left({y}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right.}\)
Substitute in the equation,
\(\displaystyle{\left({s}^{{2}}{L}{\left({y}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right)}+{L}{\left({y}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}\)
We begin with finding the Laplace for RHS,
\(\displaystyle{L}{\left({f{{\left({t}\right)}}}\right)}={\int_{{0}}^{{{3}\pi}}}{e}^{{-{s}{t}}}{\left({0}\right)}{\left.{d}{t}\right.}+{\int_{{{3}\pi}}^{{\infty}}}{e}^{{-{s}{t}}}{\left({1}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{{3}\pi}}^{{\infty}}}{e}^{{-{s}{t}}}{\left({1}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={\frac{{{e}^{{-{s}{t}}}}}{{-{s}}}}{b}{i}{g}{g}{{\mid}_{{{3}\pi}}^{{\infty}}}\)
\(\displaystyle={\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}}}}\)
Now, use the conditions given,
\(\displaystyle{s}^{{2}}{L}{\left({y}\right)}-{1}+{L}{\left({y}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}\)
\(\displaystyle{\left({s}^{{2}}+{1}\right)}{L}{\left({y}\right)}-{1}={\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}}}}\)
\(\displaystyle{\left({s}^{{2}}+{1}\right)}{L}{\left({y}\right)}={\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}}}}+{1}\)
\(\displaystyle{L}{\left({y}\right)}={\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}{\left({s}^{{2}}+{1}\right)}}}}+{\frac{{{1}}}{{{s}^{{2}}+{1}}}}\)
Now , find inverse Laplace transform
\(\displaystyle{y}{\left({t}\right)}={L}^{{-{1}}}{\left({\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}{\left({s}^{{2}}+{1}\right)}}}}\right)}+{L}^{{-{1}}}{\left({\frac{{{1}}}{{{s}^{{2}}+{1}}}}\right)}\)
Note:
\(L^{-1}\left\{F(s)\right\}=f(t) \text{ then } L^{-1}\left\{e^{-as}F(s)\right\}=H(t-a)f(t-a)\) Where H is Heaviside step function
\(\displaystyle={H}{\left({t}-{3}\pi\right)}{\left({H}{\left({t}-{3}\pi\right)}-{\cos{{\left({t}-{3}\pi\right)}}}\right)}+{\sin{{t}}}\)
Therefore,
\(\displaystyle{y}{\left({t}\right)}={H}{\left({t}-{3}\pi\right)}{\left({H}{\left({t}-{3}\pi\right)}-{\cos{{\left({t}-{3}\pi\right)}}}\right)}+{\sin{{t}}}\)

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