Question

Solve the boundary value problem

y^''+2y= −x

y(0)=0, y(1)+y^'(1)=0

Answer 1

The general solution can be written as $y=y_h+y_p.$

Homogeneous differential equation

$$y''+2y=0$$

The characteristic polynomial 

$$r^2+2=0$$

$$r_1=-i\sqrt{2}, r_2=i\sqrt{2}$$

$$y_h=c_1\cos (\sqrt{2}x)+c_2\sin (\sqrt{2}x)$$

Let $y_p=Ax+B.$ Then $y_p''=0$

$$2(Ax+B)=-x$$

$$A=-\dfrac{1}{2}, B=0$$

$$y_p=-\dfrac{1}{2}x$$

$$y=c_1\cos (\sqrt{2}x)+c_2\sin (\sqrt{2}x)-\dfrac{1}{2}x$$

$$y(0)=0:c_1\cos (\sqrt{2}(0))+c_2\sin (\sqrt{2}(0))-\dfrac{1}{2}(0)=0$$

$$c_1=0, y=c_2\sin (\sqrt{2}x)-\dfrac{1}{2}x$$

$$y'=c_2\cos (\sqrt{2}x)-\dfrac{1}{2}$$

$$y(1)+y'(1)=0$$

$$c_2\sin (\sqrt{2}(1))-\dfrac{1}{2}(1)+c_2\cos (\sqrt{2}(1))-\dfrac{1}{2}=0$$

$$c_2=\dfrac{1}{\sin(\sqrt{2})+\cos(\sqrt{2})}$$

Then

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