Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{,}\mathbf{\hspace{1em}}{\mathbf{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{6}$

Short answer

The given differential equation's solution is$\mathrm{y}=\frac{1}{6}\mathrm{t}\sin 2\mathrm{t}+2\sin 3\mathrm{t}$ .

Step-by-step solution

Given information from question

The differential equation is${\mathrm{y}}^{\text{'}}\text{'}+9\mathrm{y}=\cos 3\mathrm{t},\hspace{1em}{\mathrm{y}}_0=0,{\mathrm{y}}_0^{\text{'}}=6.$ .

Laplace transform

Application of Laplace transform:

It's used to simplify complicated differential equations by using polynomials. It's used to convert derivatives into numerous domain variables, then utilise the Inverse Laplace transform to convert the polynomials back to the differential equation.

Take the Laplace of given equation

The given equation is

$y"+9y=cos3t$

Take the Laplace of equation

$$\begin{gathered} \mathrm{L}\left\{{\mathrm{y}}^{\text{'}}\text{'}+9\mathrm{y}\right\}=\mathrm{L}\left\{\cos 3\mathrm{t}\right\} \\ \mathrm{L}\left\{\mathrm{y}"\right\}+9\mathrm{L}\left\{\mathrm{y}\right\}=\mathrm{L}\left\{\sin \mathrm{t}\right\} \\ {\mathrm{p}}^2\mathrm{L}\left\{\mathrm{y}\right\}-{\mathrm{py}}_{\mathrm{o}}-\mathrm{y}{\text{'}}_0+9\mathrm{L}\left\{\mathrm{y}\right\}=\frac{\mathrm{p}}{{\mathrm{p}}^2+9} \\ \left({\mathrm{p}}^2+9\right)\mathrm{L}\left\{\mathrm{y}\right\}-{\mathrm{py}}_{\mathrm{o}}-\mathrm{y}{\text{'}}_0=\frac{\mathrm{p}}{{\mathrm{p}}^2+9} \end{gathered}$$

Further solve

$$\begin{gathered} \left(p^2+9\right)L\left\{y\right\}-p\left(0\right)-6=\frac{p}{p^2+9} \\ \left(p^2+9\right)L\left\{y\right\}=\frac{p}{p^2+9}+6 \\ L\left\{y\right\}=\frac{p}{\left(p^2+9\right){\left(p^2+9\right)}^2}+\frac{6}{\left(p^2+9\right)} \\ L\left\{y\right\}=\frac{p}{{\left(p^2+3^2\right)}^2}+\frac{6}{\left(p^2+3^2\right)} \end{gathered}$$

The inverse Laplace is

$$\begin{gathered} y=L^{-1}\left\{\frac{p}{{\left(p^2+3^2\right)}^2}\right\}+L^{-1}\left\{\frac{6}{\left(p^2+3^2\right)}\right\} \\ y=\frac{1}{6}t\sin 2t+\frac{6}{3}\sin 3t \\ y=\frac{1}{6}t\sin 2t+2\sin 3t \end{gathered}$$

Thus, the given differential equation's solution is $y=\frac{1}{6}t\sin 2t+2\sin 3t$.