i am stumped: y" + y = -2, y(pi/8)= 1/2, y'(pi/8)

pythagorean314159

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I have been studying differential equations, and I have hit a wall on a problem. It is a boundary value problem of second order.

y" + y = -2, y(pi/8)= 1/2, y'(pi/8) = 2

Can anyone show me where to begin?
 
It is nonhomogenous differential equation (NH) with constant coefficients (\(\displaystyle y''+p(x)y'+q(x)y=g(x)\)).
The general solution is \(\displaystyle y=y_h+y_p\), where \(\displaystyle y_h\) - general solution of the homogeneous associated equation (H), and \(\displaystyle y_p\) - particular solution of the equation (NH).

(H): \(\displaystyle y''+y=0\)
Characteristic equation: \(\displaystyle r^2+1=0\), its solution \(\displaystyle r=\alpha+i\beta=i\)
\(\displaystyle y_h=c_1y_1+c_2y_2\), where \(\displaystyle y_1=e^{\alpha x}\cos{(\beta x)}=\cos x\), \(\displaystyle y_2=e^{\alpha x}\sin{(\beta x)}=\sin x\)
\(\displaystyle y_h=c_1\cos x+c_2\sin x\)

To find \(\displaystyle y_p\) we will use method of variation of parameters:
\(\displaystyle y_p=u_1(x)y_1(x)+u_2(x)y_2(x)\)

\(\displaystyle \left\{\begin{matrix}
u_1'y_1+u_2'y_2=0\\
u_1'y_1'+u_2'y_2'=g(x)
\end{matrix}\right.\)
\(\displaystyle \left\{\begin{matrix}
u_1'\cos x+u_2'\sin x=0\\
-u_1'\sin x+u_2'\cos x=-2
\end{matrix}\right.\)

\(\displaystyle u_1'=2\sin x\), \(\displaystyle u_2'=-2\cos x\)
\(\displaystyle u_1=-2\cos x\), \(\displaystyle u_2=-2\sin x\)
\(\displaystyle y_p=-2\)

\(\displaystyle y=c_1\cos x+c_2\sin x-2\)
 
By the way, this is not a "boundary value problem", it is an "initial value problem" because we are given values of y and y' at the same value of x.

That is more than just a "cosmetic" difference- whether or not a solution to an initial value problem exists and whether or not that solution is unique depends only on the equation. Whether or not a boundary value problem has a unique solution depends on the equation and the boundary conditions.
 
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