# 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

Solve the following initial value problems using Laplace Transforms:
$\frac{{d}^{2}y}{{dx}^{2}}+25y=t$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=0.04$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

FieniChoonin
Step 1
The given IVP is given as follows.

Apply the laplace transform on both sides of the differential equation as follows.
$L\left\{y{}^{″}\right\}+25L\left\{y\right\}=L\left\{t\right\}$
${s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)+25L\left\{y\right\}=\frac{1}{{s}^{2}}$
${s}^{2}L\left\{y\right\}+25L\left\{y\right\}=\frac{1}{{s}^{2}}+0.04$
$L\left\{y\right\}=\frac{1}{{s}^{2}\left({s}^{2}+25\right)}+\frac{0.04}{{s}^{2}+25}$
$L\left\{y\right\}=\frac{1}{25{s}^{2}}-\frac{1}{25\left({s}^{2}+25\right)}+\frac{0.04}{{s}^{2}+25}$
Step 2
Apply inverse laplace transforms on both sides.
$y\left(t\right)={L}^{-1}\left\{\frac{1}{25{s}^{2}}\right\}-{L}^{-1}\left\{\frac{1}{25\left({s}^{2}+25\right)}\right\}+{L}^{-1}\left\{\frac{0.04}{{s}^{2}+25}\right\}$
$=\frac{1}{25}{L}^{-1}\left\{\frac{1}{{s}^{2}}\right\}-\frac{1}{125}{L}^{-1}\left\{\frac{5}{{s}^{2}+{5}^{2}}\right\}+0.008{L}^{-1}\left\{\frac{5}{{s}^{2}+52}\right\}$
$=\frac{t}{25}-\frac{1}{125}\mathrm{sin}\left(5t\right)+0.008\mathrm{sin}\left(5t\right)$
$=\frac{t}{25}$