General Solution of 2nd Order Differential Equaiton

[mm391]

Homework Statement

Find the general solution to d²y/dx² +4y=cos(2x)

Homework Equations

The Attempt at a Solution

I have woked out what I think is the Complementary function C₁sin(2x)+C₂cos(2x) the reason it is cos and sin is because the roots are 2i and therefore the exponential and imaginary number turn it into a cos or sin.

Particular Integral:
y = a cos(2x) + b sin(2x)
y' = -2a sin(2x) + 2b cos(2x)
y'' = 4a cos(2x) - 4b sin(2x)

∴ -4a cos(2x) + 4b sin(2x) + 4a cos(2x) - 4b sin(2x) = cos(2x)

but it all cancels out to give 0=cos(2x) which surely can't be right. Have I been staring at this so long that I cannot see the obvious answer?

 

[LCKurtz]

Homework Statement

Find the general solution to d²y/dx² +4y=cos(2x)

Homework Equations

The Attempt at a Solution

I have woked out what I think is the Complementary function C₁sin(2x)+C₂cos(2x) the reason it is cos and sin is because the roots are 2i and therefore the exponential and imaginary number turn it into a cos or sin.

Particular Integral:
y = a cos(2x) + b sin(2x)
y' = -2a sin(2x) + 2b cos(2x)
y'' = 4a cos(2x) - 4b sin(2x)

∴ -4a cos(2x) + 4b sin(2x) + 4a cos(2x) - 4b sin(2x) = cos(2x)

but it all cancels out to give 0=cos(2x) which surely can't be right. Have I been staring at this so long that I cannot see the obvious answer?

Since cos(2x) and sin(2x) satisfy the homogeneous equation, of course when you plug any linear combination of them into it you are going to get zero. So they can't make the solution of the non-homogeneous equation. Your text should have a section about what to do when the right hand side is a solution of the homogeneous equation. Try$$
y_p = Cx\cos(2x)+Dx\sin(2x)$$