Question

Solve the following IVP using Laplace Transformy′′+3y′+2y=e^(-t), y(0)=0 y′(0)=0

Laplace transform
ANSWERED
asked 2021-03-09

Solve the following IVP using Laplace Transform
\(y′′+3y′+2y=e^{-t}, y(0)=0 y′(0)=0\)

Expert Answers (1)

2021-03-10

Step 1
We have to solve given IVP using Laplace transform.
Step 2
We have given
\(y′′+3y′+2y=e^{-t}, y(0)=0 y′(0)=0\)
Taking Laplace transform on given differential equation
\(\displaystyle{L}{\left({y}{''}+{3}{y}′+{2}{y}\right)}={L}{\left({e}^{{-{t}}}\right)}\)
\(\displaystyle{L}{\left({y}{''}\right)}+{3}{L}{\left({y}′\right)}+{2}{L}{\left({y}\right)}=\frac{1}{{{s}+{1}}}\ldots{\left({i}\right)}\)
We are using here
\(\displaystyle{L}{\left({y}{''}\right)}={s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}′{\left({0}\right)},{L}{\left({y}′\right)}={s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\ \text{ and }\ {L}{\left({e}^{{-{t}}}\right)}=\frac{1}{{{s}+{1}}}\)
where \(L(y)=Y(s)\)
Then from equation (i) we get,
\(\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}′{\left({0}\right)}+{3}{\left({s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right)}+{2}{Y}{\left({s}\right)}=\frac{1}{{{s}+{1}}}\)
\(\displaystyle{s}^{2}{Y}{\left({s}\right)}+{3}{s}{Y}{\left({s}\right)}+{2}{Y}{\left({s}\right)}=\frac{1}{{{s}+{1}}}\)
\(\displaystyle{\left({s}^{2}+{3}{s}+{2}\right)}{Y}{\left({s}\right)}=\frac{1}{{{s}+{1}}}\)
\(\displaystyle{\left({s}{\left({s}+{1}\right)}+{2}{\left({s}+{1}\right)}\right)}{Y}{\left({s}\right)}=\frac{1}{{{s}+{1}}}\)
\(\displaystyle{\left({s}+{1}\right)}{\left({s}+{2}\right)}=\frac{1}{{{s}+{1}}}\)
\(\displaystyle{Y}{\left({s}\right)}=\frac{1}{{{\left({s}+{1}\right)}^{2}{\left({s}+{2}\right)}}}\ldots{\left({i}{i}\right)}\)
Now
\(\displaystyle\frac{1}{{{\left({s}+{1}\right)}^{2}{\left({s}+{2}\right)}}}=\frac{A}{{{s}+{1}}}+\frac{B}{{\left({s}+{1}\right)}^{2}}+\frac{C}{{{s}+{2}}}\)
\(\displaystyle\frac{1}{{{\left({s}+{1}\right)}^{2}{\left({s}+{2}\right)}}}=\frac{{{A}{\left({s}+{1}\right)}{\left({s}+{2}\right)}+{B}{\left({s}+{2}\right)}+{C}{\left({s}+{1}\right)}^{2}}}{{{\left({s}+{1}\right)}^{2}{\left({s}+{2}\right)}}}\)
\(\displaystyle{1}={A}{\left({s}+{1}\right)}{\left({s}+{2}\right)}+{B}{\left({s}+{2}\right)}+{C}{\left({s}+{1}\right)}^{2}\)
Put \(s=-1\) we get
\(B=1\)
Put \(s=-2\) we get
\(C=1\)
Comparing \(s^2\) coefficent we get
\(A+C=0\)
\(\displaystyle\Rightarrow{A}=-{1}\)
So,
\(\displaystyle\frac{1}{{{\left({s}+{1}\right)}^{2}{\left({s}+{2}\right)}}}=-\frac{1}{{{s}+{1}}}+\frac{1}{{\left({s}+{1}\right)}^{2}}+\frac{1}{{{s}+{2}}}\)
Step 3
Now, from equation (ii)
\(\displaystyle{Y}{\left({s}\right)}=\frac{1}{{{\left({s}+{1}\right)}^{2}{\left({s}+{2}\right)}}}\)
\(\displaystyle\Rightarrow{Y}{\left({s}\right)}=-\frac{1}{{{s}+{1}}}+\frac{1}{{\left({s}+{1}\right)}^{2}}+\frac{1}{{{s}+{2}}}\)
Taking Laplace inverse on both sides we get
\(\displaystyle{L}^{ -{{1}}}{\left({Y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(\frac{{-{1}}}{{{s}+{1}}}\right)}+{L}^{ -{{1}}}{\left(\frac{1}{{\left({s}+{1}\right)}^{2}}\right)}+{L}^{ -{{1}}}{\left(\frac{1}{{{s}+{2}}}\right)}\)
\(\displaystyle\Rightarrow{y}{\left({t}\right)}=-{e}^{{-{t}}}+{t}{e}^{{-{t}}}+{e}^{{-{2}{t}}}\)
This is the solution of differential equation.

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