Trying to solve a boundary value problem

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ugetwutugiv
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Trying to solve the following boundary value problems.



y'' + 4y = cos x; y(0) = 0, y(pi) = 0
y'' + 4y = sin x; y(0) = 0, y(pi) = 0




The answer key says that there's no solution to the first part, but there is a solution to the 2nd part. I'm really lost and am not sure how to go about this. I'd greatly appreciate everyone's help on this!
 
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Were you ever taught to solve 2nd order differential equations with constant coefficients?
 
For the most part. If it were y'' + 4y = 0, I'd know what to do. But for some reason, I'm at a mental block with this. Or is this something completely different from what you're asking?
 
ugetwutugiv said:
For the most part. If it were y'' + 4y = 0, I'd know what to do. But for some reason, I'm at a mental block with this. Or is this something completely different from what you're asking?

The first step is to find the complementary solution by solving y''+4y=0. What will y be equal to for this?

You then find a particular integral for the right side.
 
y = A cos 2x + B sin 2x...that would be the complementary solution. How would I go about finding a particular integral? I was going to do an integrating factor but I think that's only for first-order ODE.
 
ugetwutugiv said:
y = A cos 2x + B sin 2x...that would be the complementary solution. How would I go about finding a particular integral? I was going to do an integrating factor but I think that's only for first-order ODE.

If the Right side is cosax,sinax, sinax+cosax, then your particular integral is Asinax+Bcosax.

To find A and B, you need to substitute back this into the differential equation and then compare coefficients.