Two-point boundary value problem

[stgermaine]

Homework Statement

Solve the given BVP or show that it has no solution. (It does have a solution)
y"+2y = x, y(0)=y(\pi)=0

Homework Equations

Characteristic polynomial is r^2 + 2 = 0. μ = √2

The Attempt at a Solution

The solution to the complementary homogeneous equation is y_h = c1 cos(√2x) + c2 sin(√2x)
Since the BVP is not homogeneous, there is a solution for the nonhomogeneous part. Let's call it y_c = d1*x + d2. Upon substituting into the problem, d1=1/2 and d2=0.

The solution is of the form y = c1 cos(√2x) + c2 sin(√2x) + (1/2)x

This was the way a similar problem was solved in the textbook. Same boundary conditions but the eqn was y"+y=x instead of y"+2y=x

The solution on the back is of the form y = c1*sin(√2x) + c2*x*sin(√2x).
Why is that?

 

[haruspex]

The solution on the back is of the form y = c1*sin(√2x) + c2*x*sin(√2x).
Why is that?

Misprint? You can easily check that this is not a solution of the equation given.