In Problems $\mathbf{1}\mathbf{-}\mathbf{14}$, solve the given initial value problem using the method of Laplace transforms

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{9}\mathbf{cost}\mathbf{+}\mathbf{7}\mathbf{sint}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{4}$

• The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
• In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Applying the Laplace transform and using its linearity we get

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}-7y\text{'}+10y\right\}=\mathcal{L}\left\{9\cos t+7\sin t\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}-7\mathcal{L}\left\{y\text{'}\right\}+10\mathcal{L}\left\{y\right\}=\frac{9s}{s^2+1}+\frac{7}{s^2+1} \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-7\left[sY\left(s\right)-y\left(0\right)\right]+10Y\left(s\right)=\frac{9s+7}{s^2+1} \\ {s}^{2}Y\left(s\right)-5s+4-7\left[sY\left(s\right)-5\right]+10Y\left(s\right) \\ =\left(s^2-7s+10\right)Y\left(s\right)=\frac{5s^3-39s^2+14s-32}{s^2+1} \\ Y\left(s\right)=\frac{5s^3-39s^2+14s-32}{\left(s^2-7s+10\right)\left(s^2+1\right)} \end{gathered}$$

Solve for the partial fraction as:

$$\begin{gathered} \frac{5s^3-39s^2+14s-32}{\left(s^2-7s+10\right)\left(s^2+1\right)}=\frac{5s^3-39s^2+14s-32}{\left(s-5\right)\left(s-2\right)\left(s^2+1\right)} \\ =\frac{A}{s-5}+\frac{B}{s-2}+\frac{Cs+D}{s^2+1} \end{gathered}$$

Solve further as:

role="math" localid="1664299653224" $$\begin{gathered} 5s^3-39s^2+14s-32=A\left(s^2+1\right)\left(s-2\right)+B\left(s^2+1\right)\left(s-5\right)+ \\ \left(Cs+D\right)\left(s-5\right)\left(s-2\right) \end{gathered}$$

Using $s=5,2,0,1$,respectively, gives

$$\begin{gathered} s=5:-312=78A\Rightarrow A=-4 \\ s=2:-120=-15B\Rightarrow B=8 \\ s=0:-32=-32+10D\Rightarrow D=0 \\ s=1:-52=4C-56\Rightarrow C=1 \end{gathered}$$

Therefore, the equation is:

$Y\left(s\right)=-\frac{4}{s-5}+\frac{8}{s-2}+\frac{s}{s^2+1}$

Using the inverse Laplace transform we obtain the solution of given differential equation.

$$\begin{gathered} y\left(t\right)=L^{-1}\left\{-\frac{4}{s-5}+\frac{8}{s-2}+\frac{s}{s^2+1}\right\}\left(t\right) \\ =-4L\left\{\frac{1}{s-5}\right\}+8L\left\{\frac{1}{s-2}\right\}+C\left\{\frac{s}{s^2+1}\right\} \\ =-4e^{5t}+8e^{2t}+\cos t \end{gathered}$$

Therefore, the solution of the initial value problem is:

$y\left(t\right)=-4e^{5t}+8e^{2t}+\cos t$

Therefore, the initial value for$y\text{'}\text{'}-7y\text{'}+10y=9\cos t+7\sin t$is$y\left(t\right)=-4e^{5t}+8e^{2t}+\cos t$