# Use Laplace transforms to solve the following initial value problem y"-y'-6y=0 y(0)=1 y'(0)=-1

Question
Laplace transform
Use Laplace transforms to solve the following initial value problem
$$y"-y'-6y=0$$
$$y(0)=1$$
$$y'(0)=-1$$

2021-02-15
Step 1
The given IVP is as follows.
$$y"-y'-6y=0$$
$$y(0)=1$$
$$y'(0)=-1$$
Apply Laplace transform on the IVP as follows.
$$y''-y'-6y=0$$
$$\displaystyle{L}{\left\lbrace{y}{''}\right\rbrace}-{L}{\left\lbrace{y}'\right\rbrace}-{L}{\left\lbrace{6}{y}\right\rbrace}={L}{\left\lbrace{0}\right\rbrace}$$
$$\displaystyle{s}^{2}{L}{\left\lbrace{y}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{s}{L}{\left\lbrace{y}\right\rbrace}+{y}{\left({0}\right)}-{6}{L}{\left\lbrace{y}\right\rbrace}={0}$$
$$\displaystyle{L}{\left\lbrace{y}\right\rbrace}{\left[{s}^{2}-{s}-{6}\right]}={s}{\left({1}\right)}+{\left(-{1}\right)}-{\left({1}\right)}$$
$$\displaystyle{L}{\left\lbrace{y}\right\rbrace}{\left[{s}^{2}-{s}-{6}\right]}={s}-{2}$$
$$\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{{{s}-{2}}}{{{s}^{2}-{s}-{6}}}$$
$$\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}$$
Step 2
Apply partial fraction on $$\displaystyle\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}$$ as follows.
$$\displaystyle\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}=\frac{A}{{{s}-{3}}}+\frac{B}{{{s}+{2}}}$$
$$s-2=A(s+2)+B(s-3)$$
$$s-2=(A+B)s+(2A-3B)$$
$$A+B=1, 2A-3B=-2$$
$$A=1-B$$
$$2(1-B)-3B=-2$$
$$-5B=-4$$
$$\displaystyle{B}=\frac{4}{{5}}$$
$$\displaystyle{A}=\frac{1}{{5}}$$
$$\displaystyle\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}=\frac{1}{{5}}{\left(\frac{1}{{{s}-{3}}}\right)}+\frac{4}{{5}}{\left(\frac{1}{{{s}+{2}}}\right)}$$
Obtain the solution of the IVP as follows.
$$\displaystyle{L}{\left\lbrace{y}\right\rbrace}={15}{\left({1}{s}-{3}\right)}+{45}{\left({1}{s}+{2}\right)}$$
Take inverse Laplace transforms as follows.
$$\displaystyle{L}^{ -{{1}}}{\left\lbrace{L}{\left\lbrace{y}\right\rbrace}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{5}}{\left(\frac{1}{{{s}-{3}}}\right)}+\frac{4}{{5}}{\left(\frac{1}{{{s}+{2}}}\right\rbrace}\right.}$$
$$\displaystyle{y}{\left({t}\right)}=\frac{1}{{5}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{3}}}\right\rbrace}+\frac{4}{{5}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}+{2}}}\right\rbrace}$$
$$\displaystyle{y}{\left({t}\right)}=\frac{1}{{5}}{e}^{{{3}{t}}}+\frac{4}{{5}}{e}^{{-{2}{t}}}$$

### Relevant Questions

Use Laplace transforms to solve the following initial value problem
$$x"-25x=9t$$
$$x(0)=x'(0)=0$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
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
Use the Laplace transform to solve the following initial value problem:
$$2y"+4y'+17y=3\cos(2t)$$
$$y(0)=y'(0)=0$$
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for $$L\left\{y(t)\right\}$$ b) Express the solution $$y(t)$$ in terms of a convolution integral
Use Laplace transform to solve the folowing initial value problem $$y"+2y'+2y=0$$
$$y(0)=2$$
$$y'(0)=-1$$
use the Laplace transform to solve the given initial-value problem. $$y"+y'-2y=10e^{-t}, y(0)=0,y'(0)=1$$
use the Laplace transform to solve the given initial-value problem.
$$y"-3y'+2y=4 \ , \ y(0)=0 \ , \ y'(0)=1$$
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
$$\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)},\ \ \ \ {y}{\left({0}\right)}={1},\ \ \ \ {y}'{\left({0}\right)}={0}\ \ {y}{\left({t}\right)}=?$$
$${y}{''}+{2}{y}'+{y}={0},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$