Find the solution of the initial value problem given below by Laplace transform. x"+16x=cos(4t) x(0)=0 text{ and } x'(0)=1

Find the solution of the initial value problem given below by Laplace transform.
$x"+16x=\mathrm{cos}\left(4t\right)$
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Step 1
We have to first take Laplace transform of the given differential equation.
step 2
$x"+16x=\mathrm{cos}\left(4t\right)$

Taking laplace transform
$\left[{s}^{2}X\left(s\right)-0-{x}^{\prime }\left(0\right)\right]+16X\left(s\right)=\frac{s}{\left({s}^{2}+16\right)}$
$⇒X\left(s\right)\left[{s}^{2}+16\right]=\frac{s}{\left({s}^{2}+16\right)}+1$
$⇒X\left(s\right)=\frac{\left({s}^{2}+s+16\right)}{\left({s}^{2}+16{\right)}^{2}}$
Taking Laplace inverse of e $⇒X\left(s\right)=\frac{1}{\left({s}^{2}+16\right)}+\frac{s}{\left({s}^{2}+16{\right)}^{2}}$
$⇒{L}^{-1}\left\{X\left(s\right)\right\}=\frac{\mathrm{sin}4t}{4}+\frac{t\mathrm{sin}4t}{8}$
Answer: $x\left(t\right)=\left(\mathrm{sin}4t\right)/4+\left(t\mathrm{sin}4t\right)/8$