Answer to Question #292623 in Differential Equations for Ability

Question

find a solution to the boundary value problem y²+4y=0 y(π/8)=0, y(π/6)=1 if the general solution to the differential equation is y(x)=c1sin2x + c2cos2x

Answer 1

$$y''+4y=0$$

The general solution is

$$y=c_1\sin (2x)+c_2 \cos(2x)$$

Given $y(\pi/8)=0$

$$0=c_1\sin (2(\pi/8))+c_2 \cos(2(\pi/8))$$

$$c_1+c_2=0$$

$$c_1=-c_2$$

Given $y(\pi/6)=1$

$$1=c_1\sin (2(\pi/6))+c_2 \cos(2(\pi/6))$$

$$c_1(\dfrac{\sqrt{3}}{2})+c_2(\dfrac{1}{2})=1$$

$$\sqrt{3}c_1+c_2=2$$

Then

$$c_2=-c_1$$

$$\sqrt{3}c_1-c_1=2$$

$$c_1=\dfrac{2}{\sqrt{3}-1}$$

$$c_1=\sqrt{3}+1$$

$$c_2=-(\sqrt{3}+1)$$

The solution to the boundary value problem is

$$y=(\sqrt{3}+1)\sin(2x)-(\sqrt{3}+1)\cos(2x)$$

$$y=(\sqrt{3}+1)\big(\sin(2x)-\cos (2x)\big)$$