Question

# Use Laplace transform to solve the following initial-value problem y"+2y'+y=0 y(0)=1, y'(0)=1 a) displaystyle{e}^{{-{t}}}+{t}{e}^{{-{t}}} b) displayst

Laplace transform

Use Laplace transform to solve the following initial-value problem
$$y"+2y'+y=0$$
$$y(0)=1, y'(0)=1$$
a) $$\displaystyle{e}^{{-{t}}}+{t}{e}^{{-{t}}}$$
b) $$\displaystyle{e}^{t}+{2}{t}{e}^{t}$$
c) $$\displaystyle{e}^{{-{t}}}+{2}{t}{e}^{t}$$
d) $$\displaystyle{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}$$
e) $$\displaystyle{2}{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}$$
f) Non of the above

2021-02-20
Step 1
Given initial value problem,
$$y''+2y'+y=0$$
$$y(0)=1$$
$$y'(0)=1$$
Step 2
Taking inverse Laplace transform,
$$\displaystyle{L}{\left[{y}{''}+{2}{y}'+{y}\right]}={0}$$
$$\displaystyle{L}{\left[{y}{''}\right]}+{2}{L}{\left[{y}'\right]}+{L}{\left[{y}\right]}={0}$$
Use the formula such that,
$$\displaystyle{L}{\left[{y}{''}\right]}={s}^{2}{L}{\left[{y}\right]}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}$$
$$\displaystyle{L}{\left[{y}'\right]}={s}{L}{\left[{y}\right]}-{y}{\left({0}\right)}$$
Then,
$$\displaystyle{s}^{2}{L}{\left[{y}\right]}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{2}{\left[{s}{L}{\left[{y}\right]}-{y}{\left({0}\right)}\right]}+{L}{\left[{y}\right]}={0}$$
$$\displaystyle{s}^{2}{L}{\left[{y}\right]}-{s}-{1}+{2}{\left[{s}{L}{\left[{y}\right]}-{1}\right]}+{L}{\left[{y}\right]}={0}$$
$$\displaystyle{s}^{2}{L}{\left[{y}\right]}-{s}-{1}+{2}{s}{L}{\left[{y}\right]}-{2}+{L}{\left[{y}\right]}={0}$$
$$\displaystyle{\left({s}^{2}+{2}{s}+{1}\right)}{L}{\left[{y}\right]}-{s}-{3}={0}$$
$$\displaystyle{L}{\left[{y}\right]}=\frac{{{s}+{3}}}{{{s}^{2}+{2}{s}+{1}}}$$
Step 3
Taking inverse Laplace transform of both sides,
$$\displaystyle{y}={L}^{ -{{1}}}{\left[\frac{{{s}+{3}}}{{{s}^{2}+{2}{s}+{1}}}\right]}$$
$$\displaystyle={L}^{ -{{1}}}{\left[\frac{{{s}+{1}}}{{\left({s}+{1}\right)}^{2}}+\frac{2}{{\left({s}+{1}\right)}^{2}}\right]}$$
$$\displaystyle={L}^{ -{{1}}}{\left[\frac{1}{{{s}+{1}}}\right]}+{2}{L}^{ -{{1}}}{\left[\frac{1}{{\left({s}+{1}\right)}^{2}}\right]}$$
$$\displaystyle={e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}$$
Step 4
Hence, the solution of given initial value problem is
$$\displaystyle{y}{\left({t}\right)}={e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}$$