Use Laplace transforms to solve the initial value problems.y"+4y=2\delta(t-\pi), y(0)=1 , y'(0)=0

Phoebe 2021-09-16 Answered

Use Laplace transforms to solve the initial value problems.
\(y"+4y=2\delta(t-\pi), y(0)=1 , y'(0)=0\)

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

AGRFTr
Answered 2021-09-17 Author has 12458 answers

We know that \(L\left\{\delta(t-t_0)\right\}=e^{-s t_0}\)
and \(L^{-1}\left\{e^{-as} F(s)\right\}=f(t-a)u(t-a)\)
Now , given that, \(y"+4y=2\delta(t-\pi); y(0)=1 . y'(0)=0\)
Taking Laplace transformation on both sides we get,
\(L\left\{y"+4y\right\}=2L\left\{\delta(t-\pi)\right\}\)
\(\Rightarrow L\left\{y"\right\}+4L\left\{y\right\}=2L\left\{\delta(t-\pi)\right\}\)
\(\Rightarrow sY(s)-sy(0)-y'(0)+4y(s)=2e^{-\pi s}\)
\(\Rightarrow s^2Y(s)-s-0+4y(s)=2e^{-\pi s} \left[\because y(s)=L\left\{y\right\}\right]\)
\(\displaystyle\Rightarrow{y}{\left({s}\right)}{\left[{s}^{{2}}+{4}\right]}={2}{e}^{{-\pi{s}}}+{s}\)
\(\displaystyle\Rightarrow{y}{\left({s}\right)}={2}{\frac{{{e}^{{-\pi{s}}}}}{{{s}^{{2}}+{4}}}}+{\frac{{{s}}}{{{s}^{{2}}+{4}}}}\)
Now taking inverse Laplace transformation,
\(y(d)=L^{-1}\left\{\frac{2e^{-\pi s}}{s^2+4}\right\}+L^{-1}\left\{\frac{s}{s^2+4}\right\}\)
\(\therefore y(d)=L^{-1}\left\{e^{-\pi s} \frac{2}{s^2+2^2}\right\}+L^{-1}\left\{\frac{s}{s^2+2^2}\right\}\ \ \ (2)\)
Now, \(L^{-1}\left\{e^{-\pi s} \frac{2}{s^2+2^2}\right\}\)
Let, \(\displaystyle{F}{\left({s}\right)}={\frac{{{2}}}{{{s}^{{2}}+{2}^{{2}}}}},\ \text{ Then }\ {f{{\left({t}\right)}}}={\sin{{2}}}{t}\)
\(\displaystyle\therefore{f{{\left({t}-\pi\right)}}}={\sin{{2}}}{\left({t}-\pi\right)}\)
\(L^{-1}\left\{e^{-\pi s} \frac{2}{s^2+2^2}\right\}=f(t-2)u(t-a)\)
\(\displaystyle={\sin{{\left({2}{t}-{2}\pi\right)}}}{u}{\left({t}-\pi\right)}\)
\(\displaystyle={\sin{{2}}}{\left({t}-\pi\right)}{u}{\left({t}-\pi\right)}\)
and \(L^{-1}\left\{\frac{s}{s^2+2^2}\right\}=\cos 2t\)
From (2) we get
\(\displaystyle{y}{\left({t}\right)}={\sin{{2}}}{\left({t}-\pi\right)}{u}{\left({t}-\pi\right)}+{\cos{{2}}}{t}\)
\(\displaystyle\Rightarrow{y}{\left({t}\right)}={\sin{{2}}}{t}{u}{\left({t}-\pi\right)}+{\cos{{2}}}{t}\)

Not exactly what you’re looking for?
Ask My Question
48
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-09-18
Use Laplace transforms to solve the initial value problems.
\(\displaystyle{y}\text{}{y}'=\delta{\left({t}\right)}-\delta{\left({t}-{3}\right)},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={0}\)
asked 2021-09-06

Solve the IVP using Laplace transforms:
\(\displaystyle{}\text{}{}{y}''+{3}{y}={U}{\left({t}-\pi\right)}{e}^{{-{2}{\left({t}-\pi\right)}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}\)

asked 2021-09-22
Use Laplace transforms to solve the following initial value problem.
\(\displaystyle{x}^{{{\left({3}\right)}}}+{x}\text{-6x'=0, x(0)=0 , x'(0)=x}{\left({0}\right)}={7}\)
The solution is x(t)=?
asked 2021-09-06

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

asked 2021-09-26

Use Laplace transforms to solve the following initial value problem.
\(x''+x=3 \cos 3t , x(0)=1 , x'(0)=0\)
The solution is \(x(t)=?\)

asked 2021-03-04

use the Laplace transform to solve the given initial-value problem.
\(y"+y=f(t)\)
\(y(0)=0 , y'(0)=1\) where
\(\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}\)

asked 2021-09-16

Solve the initial value problem below using the method of Laplace transforms.
\(\displaystyle{y}\text{}{8}{y}'+{41}{y}={29}{e}^{{{2}{t}}}\)
\(y(0)=1 , y'(0)=7\)

...