# Solve the initial value problem {(y+16y=cos(4t)),(y(0)=1),(y'(0)=1):} using the Laplace transform.

Solve the initial value problem $\left\{\begin{array}{c}y+16y=\mathrm{cos}\left(4t\right)\\ y\left(0\right)=1\\ {y}^{\prime }\left(0\right)=1\end{array}$ using the Laplace transform.

You can still ask an expert for help

• 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

Obiajulu
Step 1
Given: $y+16y=\mathrm{cos}\left(4t\right),y\left(0\right)=1,{y}^{\prime }\left(0\right)=1$
Step 2
Explanation:
$y+16y=\mathrm{cos}\left(4t\right),y\left(0\right)=1,{y}^{\prime }\left(0\right)=1$
Taking Laplace transform on both sides
$L\left[y{}^{″}+16y\right]=L\left[\mathrm{cos}\left(4t\right)\right]$
$L\left(y\left(t\right)\right)+L\left(16y\left(t\right)\right)=L\left[\mathrm{cos}\left(4t\right)\right]\dots \left(i\right)$
We know that if $L\left(y\left(t\right)\right)=y\left(s\right)$
then $L\left(y{}^{″}\left(t\right)\right)={s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[\mathrm{cos}\left(at\right)\right]=\frac{s}{{s}^{2}+{a}^{2}}$
Using these in equation (i), we get
${s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+16y\left(s\right)=\frac{s}{{s}^{2}+{4}^{2}}$
Using the given initial conditions $y\left(0\right)=1,{y}^{\prime }\left(0\right)=1$ then
$\left({s}^{2}+16\right)y\left(s\right)-s\left(1\right)-1=\frac{s}{{s}^{2}+{4}^{2}}$
$\left({s}^{2}+{4}^{2}\right)y\left(s\right)=1+s+\frac{s}{{s}^{2}+{4}^{2}}$
$y\left(s\right)=\frac{1}{{s}^{2}+{4}^{2}}+\frac{s}{{s}^{2}+{4}^{2}}+\frac{s}{\left({s}^{2}+{4}^{2}\right)\left({s}^{2}+{4}^{2}\right)}$
$y\left(s\right)=\frac{1}{{s}^{2}+{4}^{2}}+\frac{s}{{s}^{2}+{4}^{2}}+\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}$
Taking inverse Laplace transform on both sides
${L}^{-1}\left(y\left(s\right)\right)={L}^{-1}\left(\frac{1}{{s}^{2}+{4}^{2}}+\frac{s}{{s}^{2}+{4}^{2}}+\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}\right)$