Question

Solve the equation with the help of laplace transform (d^2y+dx=2 ,when x(0)=3, x’(0)=1, where d=d/█(dtà)​

Answer 1

Correct Question:-

Solve the following differential equation using Laplace transformation.

`\rm{x''(t)+x'(t)=2,\quad x(0)=3,\quad x'(0)=1}`

where
`\rm{x'(t)=(dx)/(dt),\quad x''(t)=(d^2x)/(dt^2).}`

`\quad`

Solution:-

Let,

`\cal{L}\{\rm{x(t)}\}=X(s)`

`\cal{L}^{\rm{-1}}\{\rm{X(s)}\}=x(t)`

Given,

`\longrightarrow\rm{x''(t)+x'(t)=2}`

We take Laplace transformation of both sides of the equation.

`\longrightarrow\cal{L}\{\rm{x''(t)+x'(t)}\}=\cal{L}\{\rm{2}\}`

`\longrightarrow\cal{L}\{\rm{x''(t)}\}+\cal{L}\{\rm{x'(t)}\}=\cal{L}\{\rm{2}\}`

We have,

`\cal{L}\{\rm{x''(t)}\}=\rm{s^2\,X(s)-s\,x(0)-x'(0)}`

`\cal{L}\{\rm{x'(t)}\}=\rm{s\,X(s)-\,x(0)}`

`\cal{L}\{\rm{2}\}=\rm{(2)/(s)}`

Then,

`\small\text{$\longrightarrow\rm{\big[s^2\,X(s)-s\,x(0)-x'(0)\big]+\big[s\,X(s)-x(0)\big]=(2)/(s)}$}`

`\longrightarrow\rm{s(s+1)\,X(s)-(s+1)\,x(0)-x'(0)=(2)/(s)}`

Given that
`\rm{x(0)=3}` and
`\rm{x'(0)=1.}` Then,

`\longrightarrow\rm{s(s+1)\,X(s)-3(s+1)-1=(2)/(s)}`

`\longrightarrow\rm{s(s+1)\,X(s)-3s-4=(2)/(s)}`

`\longrightarrow\rm{s(s+1)\,X(s)=(2)/(s)+3s+4}`

`\longrightarrow\rm{X(s)=(2)/(s^2(s+1))+(3s+4)/(s(s+1))}`

`\longrightarrow\rm{X(s)=(3s^2+4s+2)/(s^2(s+1))\quad\dots(1)}`

We will factorise this expression by partial fractions.

Assume,

`\longrightarrow\rm{(3s^2+4s+2)/(s^2(s+1))=(As+B)/(s^2)+(C)/(s+1)}`

`\longrightarrow\rm{(3s^2+4s+2)/(s^2(s+1))=((As+B)(s+1)+Cs^2)/(s^2(s+1))}`

`\longrightarrow\rm{3s^2+4s+2=(As+B)(s+1)+Cs^2}`

`\longrightarrow\rm{3s^2+4s+2=(A+C)s^2+(A+B)s+B}`

Equating corresponding coefficients,

`\rm{A+C=3}`

`\rm{A+B=4}`

`\rm{B=2}`

Solving each equation we get,

`\rm{A=2}`

`\rm{B=2}`

`\rm{C=1}`

Therefore,

`\longrightarrow\rm{(3s^2+4s+2)/(s^2(s+1))=(2s+2)/(s^2)+(1)/(s+1)}`

`\longrightarrow\rm{(3s^2+4s+2)/(s^2(s+1))=(2)/(s)+(2)/(s^2)+(1)/(s+1)}`

Then (1) becomes,

`\longrightarrow\rm{X(s)=(2)/(s)+(2)/(s^2)+(1)/(s+1)}`

Now we will inverse Laplace transformation to obtain the solution.

`\longrightarrow\cal{L}^{\rm{-1}}\{\rm{X(s)}\}=\cal{L}^{\rm{-1}}\left\{\rm{(2)/(s)+(2)/(s^2)+(1)/(s+1)}\right\}`

`\small\text{$\longrightarrow\rm{x(t)}=\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{(1)/(s)}\right\}+\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{(1)/(s^2)}\right\}+\cal{L}^{\rm{-1}}\left\{\rm{(1)/(s+1)}\right\}$}`

`\small\text{$\longrightarrow\rm{x(t)}=\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{(1)/(s)}\right\}+\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{(1!)/(s^(1+1))}\right\}+\cal{L}^{\rm{-1}}\left\{\rm{(1)/(s-(-1))}\right\}$}`

We have,

`\cal{L}^{\rm{-1}}\left\{\rm{(1)/(s)}\right\}=\rm{1}`

`\cal{L}^{\rm{-1}}\left\{\rm{(1!)/(s^(1+1))}\right\}=\rm{t}`

`\cal{L}^{\rm{-1}}\left\{\rm{(1)/(s-(-1))}\right\}=\rm{e^(-t)=(1)/(e^t)}`

Hence,

`\longrightarrow\rm{\underline{\underline{x(t)=2+2t+(1)/(e^t)}}}`

This is the solution to our differential equation.

`\quad`

Some results of Laplace Transformation:-

`\boxed{\begin{array}c&\\\rm{x(t)}=\cal{L}^{\rm{-1}}\{\rm{X(s)}\}&\rm{X(s)}=\cal{L}\{\rm{x(t)}\}\\\\=============&==================\\\\\rm{(d^nx)/(dt^n)}&\rm{s^n\,X(s)}-\displaystyle\sum_{\rm{r=0}}^{\rm{n-1}}\rm{s^(n-r-1)\,(d^rx)/(dt^r)(t=0)}\\\\----------&--------------\\\\\rm{x'(t)}&\rm{s\,X(s)-x(0)}\\\\----------&--------------\\\\\rm{x''(t)}&\rm{s^2\,X(s)-s\,x(0)-x'(0)}\\\\----------&--------------\\\\\rm{a}&\rm{(a)/(s)}\\\\----------&--------------\\\\\rm{t^n,\ n}\in\mathbb{N}&\rm{(n!)/(s^(n+1))}\\\\----------&--------------\\\\\rm{e^(at)}&\rm{(1)/(s-a)}\\&\end{array}}`