Question

Solve the given initial-value problem in which the input function $\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is discontinuous. [Hint: Solve each problem on two intervals, and then find a solution so that y and y^' are continuous at $\mathbf{x}\mathbf{=}\mathbf{\pi }\mathbf{/}\mathbf{2}$and at $\mathbf{x}\mathbf{=}\mathbf{\pi }\mathbf{]}$

$\mathbf{y}\mathbf{"}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{where}\mathbf{}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{sinx}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{\pi }\mathbf{/}\mathbf{2}\\ \mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{>}\mathbf{\pi }\mathbf{/}\mathbf{2}\end{array}\mathbf{}\mathbf{}$

Short answer

The General solution for given initial-value

$y\left(x\right)=\left\{\begin{array}{l}\cos \left(2x\right)+\frac{5}{6}\sin \left(2x\right)+\frac{1}{3}\sin \left(x\right),0\le x\le \mathrm{\pi }/2\\ \frac{2}{3}\cos \left(2x\right)+\frac{5}{6}\sin \left(2x\right),x>\mathrm{\pi }/2\end{array}\right.$

Step-by-step solution

Determine the problem with input function g(x)

Solve the auxiliary equation to get y_h

_{$$\begin{gathered} {\lambda }^2+4=0\Rightarrow \lambda =\pm 2i,y_h=c_1\cos \left(2x\right)+c_2\sin \left(2x\right) \\ y"+4y= \\ \left\{\begin{array}{l}\sin \left(x\right),0\le x\le \mathrm{\pi }/2\\ 0,x>\mathrm{\pi }/2\end{array}\right. \end{gathered}$$}

For the upper definition with g(x) = sin (x),y_p= A cos(x)+B sin(x)

y" = -A cos(x) -B sin (x)

Substitute into the equation, we get

-A cos(x) -B Sin(x) + 4A cos(x) + 4B sin (x) = 3A Cos (x) + 3B sin (x) = sin (x)

Comparing coefficients in both sides

$\Rightarrow A=0,B=\frac{1}{3},\Rightarrow y_p=\frac{1}{3}\sin \left(x\right)$

The general solution $\mathrm{y}={\mathrm{y}}_{\mathrm{p}}+{\mathrm{y}}_{\mathrm{h}}={\mathrm{c}}_1\cos \left(2\mathrm{x}\right)+{\mathrm{c}}_2\sin \left(2\mathrm{x}\right)+\frac{1}{3}\sin \left(\mathrm{x}\right)$

Determine the problem with two intervals

Apply $\mathrm{y}\left(0\right)=1,{\mathrm{c}}_{,}\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)+\frac{1}{3}\sin \left(0\right)=1\Rightarrow {\mathrm{c}}_1=1$

Apply $\mathrm{y}\left(0\right)=2,-2{\mathrm{c}}_1\sin \left(0\right)+2{\mathrm{c}}_2\cos \left(0\right)+\frac{1}{3}\cos \left(0\right)=2\Rightarrow {\mathrm{c}}_2=\frac{5}{6}$

$\mathrm{y}=\cos \left(2\mathrm{x}\right)+\frac{5}{6}\sin \left(2\mathrm{x}\right)+\frac{1}{3}\sin \left(\mathrm{x}\right)$

For the upper definition with g(x) = 0, y = y_h = c₃ cos(2x) + c₄ sin(2x)

Sustain Continuity at $\mathrm{x}=\frac{\mathrm{\pi }}{2}$

role="math" localid="1667814980779" $$\begin{gathered} {\mathrm{y}}_{\mathrm{upper}}\left(\frac{\mathrm{\pi }}{2}\right)=-1+\frac{1}{3}=\frac{-2}{3} \\ {\mathrm{y}}_{\mathrm{lower}}\left(\frac{\mathrm{\pi }}{2}\right)=-{\mathrm{c}}_3\Rightarrow {\mathrm{c}}_3=\frac{2}{3} \end{gathered}$$

role="math" localid="1667814994093" $$\begin{gathered} \mathrm{y}{\text{'}}_{\mathrm{upper}}\left(\frac{\mathrm{\pi }}{2}\right)=\frac{-5}{3} \\ \mathrm{y}{\text{'}}_{\mathrm{lower}}\left(\frac{\mathrm{\pi }}{2}\right)=-2{\mathrm{c}}_4\Rightarrow {\mathrm{c}}_4=\frac{5}{6} \end{gathered}$$

$\mathrm{y}=\frac{2}{3}\cos \left(2\mathrm{x}\right)+\frac{5}{6}\sin \left(2\mathrm{x}\right)$

The general solution

$\mathrm{y}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\cos \left(2\mathrm{x}\right)+\frac{5}{6}\sin \left(2\mathrm{x}\right)+\frac{1}{3}\sin \left(\mathrm{x}\right),0\le \mathrm{x}\le \mathrm{\pi }/2\\ \frac{2}{3}\cos \left(2\mathrm{x}\right)+\frac{5}{6}\sin \left(2\mathrm{x}\right),\mathrm{x}>\mathrm{\pi }/2\end{array}\right.$