Question

In Problems $\mathbf{7}\mathbf{-}\mathbf{10}\mathbf{,}\mathit{x}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$is a two-parameter family of solutions of the second order${\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{x}\mathbf{=}\mathbf{0}$. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

$\mathit{x}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{4}\mathbf{)}\mathbf{=}\sqrt{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{4}\mathbf{)}\mathbf{=}\mathbf{2}\sqrt{\mathbf{2}}$

Short answer

The solution is$x=-\cos t+3\sin t$.

Step-by-step solution

Definition:

An initial value problem is a differential equation with some initial conditions.

Differentiate & evaluate :

Given$x=c_1\cos t+c_2\sin t$

Differentiating we obtain$x^{\text{'}}=-c_1\sin t+c_2\cos t$

Now applying initial condition i.e; $x\left(\frac{\pi }{4}\right)=\sqrt{2}$we have:

$$\begin{gathered} \sqrt{2}=c_1\cos \frac{\pi }{4}+c_2\sin \frac{\pi }{4} \\ =c_1\left(\frac{1}{\sqrt{2}}\right)+c_2\left(\frac{1}{\sqrt{2}}\right) \\ =\frac{c_1+c_2}{\sqrt{2}} \end{gathered}$$

$\Rightarrow c_1+c_2=2----\left(1\right)$

Apply second condition:

Also, $x^{\text{'}}\left(\frac{\pi }{4}\right)=2\sqrt{2}$

$\Rightarrow 2\sqrt{2}=-c_1\sin \frac{\pi }{4}+c_2\cos \frac{\pi }{4}$

$=\frac{-c_1+c_2}{\sqrt{2}}\Rightarrow -c_1+c_2=4----\left(2\right)$

Solve for constants:

Add equation (1) & (2) to get:

$2c_2=6$

$\therefore c_2=3$

Equation (1) gives :

$c_1=2-c_2$

$$\begin{gathered} =2-3 \\ =-1 \end{gathered}$$

Therefore the solution of given differential equation is$x=-\cos t+3\sin t$ .