# Solve the following initial value problems using Laplace Transforms: displaystylefrac{{{d}^{2}{y}}}{{{left.{d}{x}right.}^{2}}}+{25}{y}={t} y(0)=0 y'(0)=0.04

Question
Laplace transform
Solve the following initial value problems using Laplace Transforms:
$$\displaystyle\frac{{{d}^{2}{y}}}{{{\left.{d}{x}\right.}^{2}}}+{25}{y}={t}$$
$$y(0)=0$$
$$y'(0)=0.04$$

2021-02-27
Step 1
The given IVP is given as follows.
$$\displaystyle\frac{{{d}^{2}{y}}}{{{\left.{d}{x}\right.}^{2}}}+{25}{y}={t},\ \ \ {y}{\left({0}\right)}={0},\ \ \ {y}'{\left({0}\right)}={0.04}$$
Apply the laplace transform on both sides of the differential equation as follows.
$$\displaystyle{L}{\left\lbrace{y}{''}\right\rbrace}+{25}{L}{\left\lbrace{y}\right\rbrace}={L}{\left\lbrace{t}\right\rbrace}$$
$$\displaystyle{s}^{2}{L}{\left\lbrace{y}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{25}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{s}^{2}}$$
$$\displaystyle{s}^{2}{L}{\left\lbrace{y}\right\rbrace}+{25}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{s}^{2}}+{0.04}$$
$$\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}{\left({s}^{2}+{25}\right)}}}+\frac{{{0.04}}}{{{s}^{2}+{25}}}$$
$$\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{25}{s}^{2}}}-\frac{1}{{{25}{\left({s}^{2}+{25}\right)}}}+\frac{{{0.04}}}{{{s}^{2}+{25}}}$$
Step 2
Apply inverse laplace transforms on both sides.
$$\displaystyle{y}{\left({t}\right)}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{25}{s}^{2}}}\right\rbrace}-{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{25}{\left({s}^{2}+{25}\right)}}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace\frac{{{0.04}}}{{{s}^{2}+{25}}}\right\rbrace}$$
$$\displaystyle=\frac{1}{{25}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{s}^{2}}\right\rbrace}-\frac{1}{{125}}{L}^{ -{{1}}}{\left\lbrace\frac{5}{{{s}^{2}+{5}^{2}}}\right\rbrace}+{0.008}{L}^{ -{{1}}}{\left\lbrace\frac{5}{{{s}^{2}+{52}}}\right\rbrace}$$
$$\displaystyle=\frac{t}{{25}}-\frac{1}{{125}} \sin{{\left({5}{t}\right)}}+{0.008} \sin{{\left({5}{t}\right)}}$$
$$\displaystyle=\frac{t}{{25}}$$

### Relevant Questions

Use the table of Laplace transform and properties to obtain the Laplace transform of the following functions. Specify which transform pair or property is used and write in the simplest form.
a) $$x(t)=\cos(3t)$$
b)$$y(t)=t \cos(3t)$$
c) $$z(t)=e^{-2t}\left[t \cos (3t)\right]$$
d) $$x(t)=3 \cos(2t)+5 \sin(8t)$$
e) $$y(t)=t^3+3t^2$$
f) $$z(t)=t^4e^{-2t}$$
Use the Laplace transform to solve the given initial-value problem.
$$dy/dt-y=z,\ y(0)=0$$
Solve the following initial value problems using Laplace Transforms:
$$a) \frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y=1,$$
$$y(0=0) , y'(0)=0$$
$$b) \frac{d^2y}{dt^2}+4\frac{dy}{dt}=1,$$
$$y(0=0) , y'(0)=0$$
$$c) 2\frac{d^2y}{dt^2}+3\frac{dy}{dt}-2y=te^{-2t},$$
$$y(0=0) , y'(0)=-2$$
Find the Laplace transform of the function $$L\left\{f^{(9)}(t)\right\}$$
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 by using Laplace Transform and explain the steps in brief
$$\displaystyle\frac{{{d}^{2}{x}}}{{{\left.{d}{t}\right.}^{2}}}+{2}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}+{x}={3}{t}{e}^{{-{t}}}$$
Given $$x(0)=4, x'(0)=2$$
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.
Using Laplace Transform , solve the following differential equation
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
$$y"-4y'=-4te^{2t}, y_0=0, y'_0 =1$$
$$y"+2y'+10y=-6e^{-t}\sin3t, y_0=0, y'_0=1$$