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ugetwutugiv
Nov16-08, 12:45 AM
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!!!

rock.freak667
Nov16-08, 12:46 AM
Were you ever taught to solve 2nd order differential equations with constant coefficients?

ugetwutugiv
Nov16-08, 12:48 AM
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?

rock.freak667
Nov16-08, 12:49 AM
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.

ugetwutugiv
Nov16-08, 12:56 AM
y = A cos 2x + B sin 2x.....that would be the complementary solution. How would I go about finding a particular integral? I was gonna do an integrating factor but I think that's only for first-order ODE.

rock.freak667
Nov16-08, 12:59 AM
y = A cos 2x + B sin 2x.....that would be the complementary solution. How would I go about finding a particular integral? I was gonna 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.