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

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

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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