# Use the Laplace transform to solve the given initial-value problem. displaystyle{y}{''}-{6}{y}'+{13}{y}={0}, {y}{left({0}right)}={0}, {y}'{left({0}right)}=-{9}

Question
Laplace transform
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}-{6}{y}'+{13}{y}={0},\ \ \ {y}{\left({0}\right)}={0},\ \ \ {y}'{\left({0}\right)}=-{9}$$

2020-10-29
Step 1
The given differential equation is
$$y''-6y'+13y=0$$
Step 2
Simplify the initial value problem using Laplace transform:
$$\displaystyle{L}{\left({y}\text{}-{6}{y}'+{13}{y}\right)}={L}{\left({0}\right)}$$
$$\displaystyle{L}{\left({y}\text{}\right)}-{6}{L}{\left({y}'\right)}+{13}{L}{\left({y}\right)}={0}$$
$$\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{6}{\left({s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right)}+{13}{Y}{\left({s}\right)}={0}$$
$$\displaystyle{s}^{2}{Y}{\left({s}\right)}+{9}-{6}{s}{Y}{\left({s}\right)}+{13}{Y}{\left({s}\right)}={0}$$
$$\displaystyle{Y}{\left({s}\right)}=-\frac{9}{{{s}^{2}-{6}{s}+{9}+{4}}}$$
Step 3
Simplify further,
$$\displaystyle{Y}{\left({s}\right)}=-\frac{9}{{{\left({s}-{3}\right)}^{2}+{4}}}$$
$$\displaystyle{Y}{\left({s}\right)}=-\frac{9}{{{\left({s}-{3}\right)}^{2}+{2}^{2}}}$$
Take inverse laplace on both sides,
$$\displaystyle{L}^{ -{{1}}}{\left({Y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(-\frac{9}{{{\left({s}-{3}\right)}^{2}+{2}^{2}}}\right)}$$
$$\displaystyle{y}{\left({t}\right)}=-\frac{9}{{2}}{L}^{ -{{1}}}{\left(\frac{2}{{{\left({s}-{3}\right)}^{2}+{2}^{2}}}\right)}$$
$$\displaystyle{y}{\left({t}\right)}=-\frac{9}{{2}}{e}^{{{3}{t}}} \sin{{2}}{t}$$
Hence , the solution of the initial value problem is $$\displaystyle{y}{\left({t}\right)}=-\frac{9}{{2}}{e}^{{{3}{t}}} \sin{{2}}{t} ### Relevant Questions asked 2021-03-04 use the Laplace transform to solve the given initial-value problem. \(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.}$$
Use the Laplace transform to solve the given initial-value problem.
$$\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)}=?$$
Use the Laplace transform to solve the given initial-value problem
$${y}{''}+{2}{y}'+{y}={0},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Solve the initial value problem $$\displaystyle{\left\lbrace\begin{matrix}{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}}\\{y}{\left({0}\right)}={1}\\{y}'{\left({0}\right)}={1}\end{matrix}\right.}$$ using the Laplace transform.
Use the Laplace transform to solve the following initial value problem:
$$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)$$ in terms of a convolution integral
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\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}-?$$
b) Obtain the solution y(t).
y(t) - ?
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
$$y"+y=g(t) , y(0)=-4 , y'(0)=0$$
where $$g{{\left({t}\right)}}={\left\lbrace\begin{matrix}{t}&{t}<{4}\\{5}&{t}>{4}\end{matrix}\right.}$$
Y(s)-?
Use Laplace transform to solve the following initial-value problem
$$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
$${y}\text{}{\left({t}\right)}-{y}'{\left({t}\right)}={e}^{t} \cos{{\left({t}\right)}}+ \sin{{\left({t}\right)}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}$$
use the Laplace transform to solve the given initial-value problem. $$y"+y'-2y=10e^{-t}, y(0)=0,y'(0)=1$$