Question

In Problemssolve the given initial-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\left(\frac{\pi }{8}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(\frac{\pi }{8}\right)\mathbf{=}\mathbf{2}$

Short answer

The solution is

$y\left(x\right)=\sqrt{2}\sin 2x-\frac{1}{2}$

Step-by-step solution

Complementary function

Consider the differential equation$y\text{'}\text{'}+4y=-2,\text{y}\left(\frac{\pi }{8}\right)=\frac{1}{2},\text{y}\text{'}\left(\frac{\pi }{8}\right)=2\dots \dots \left(1\right)$

The auxiliary equation corresponding to homogenous equation gives:

$$\begin{gathered} m^2+4=0 \\ m=\pm 2i \end{gathered}$$

The complementary function is$y_c=c_1\cos 2x+c_2\sin 2x$

$y_c=c_1\cos 2x+c_2\sin 2x$

General solution

Let particular solution be of the form, $y_p=A$

Differentiating this expression and substituting the results into the differential equation,

$$\begin{gathered} y\text{'}\text{'}+4y=-2 \\ {y}_p\text{'}\text{'}+4y_p=-2 \\ 0+4A=-2\text{Since}{y}_p\text{'}\text{'}=0 \\ A=-\frac{2}{4} \\ A=-\frac{1}{2} \end{gathered}$$

Therefore, the particular solution $y_p=-\frac{1}{2}$

Thus, the general solution of the given differential equation is, $y=y_c+y_p$

The general solution is,

$y\left(x\right)=C_1\cos 2x+C_2\sin 2x-\frac{1}{2} ······\left(2\right)$

Apply the first initial condition

Differentiate equation (2).

$y\text{'}\left(x\right)=-2c_1\sin 2x+2c_2\cos 2x······\left(3\right)$

Apply the initial condition, $y\left(\frac{\pi }{8}\right)=\frac{1}{2}$and$y\text{'}\left(\frac{\pi }{8}\right)=2$

Substitute $x=\frac{\pi }{8}$and $y=\frac{1}{2}$in the equation (2):

$$\begin{gathered} y\left(x\right)=c_1\cos 2x+c_2\sin 2x-\frac{1}{2} \\ y\left(\frac{\pi }{8}\right)=c_1\cos 2\left(\frac{\pi }{8}\right)+c_2\sin 2\left(\frac{\pi }{8}\right)-\frac{1}{2} \\ \frac{1}{2}=c_1\cos \left(\frac{\pi }{4}\right)+c_2\sin \left(\frac{\pi }{4}\right)-\frac{1}{2} \\ \frac{c_1}{\sqrt{2}}+\frac{c_2}{\sqrt{2}}=1 \\ {c}_1+c_2=\sqrt{2}······\left(4\right) \end{gathered}$$

Apply the second initial condition

Substitute $y\text{'}\left(\frac{\pi }{8}\right)=2$in the equation (3)

role="math" $$\begin{gathered} y\text{'}\left(\frac{\pi }{8}\right)=-2c_1\sin 2\left(\frac{\pi }{8}\right)+2c_2\cos 2\left(\frac{\pi }{8}\right) \\ 2=-2c_1\sin \left(\frac{\pi }{4}\right)+2c_2\cos \left(\frac{\pi }{4}\right) \\ 2=-2\frac{c_1}{\sqrt{2}}+2\frac{c_2}{\sqrt{2}} \\ 2=-\sqrt{2}{c}_1+\sqrt{2}{c}_2 \\ -c_1+c_2=\sqrt{2}······\left(5\right) \end{gathered}$$

Find the Solution

Solving equation (4) & (5) we have:

$$\begin{gathered} 2c_2=2\sqrt{2} \\ {c}_2=\sqrt{2} \end{gathered}$$

From the equation, (4)

$$\begin{gathered} c_1=\sqrt{2}-c_2 \\ {c}_1=\sqrt{2}-\sqrt{2} \\ {c}_1=0 \end{gathered}$$

Therefore, the solution of the given initial -value problem is

$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$