# Thread: 2nd order differential complex!

1. ## 2nd order differential complex!

I need help with the first part mostly!

The Question:

$\frac { { d }^{ 2 }y }{ { d }t^{ 2 } } +y=\quad (2cosx)+\quad (2){ e }^{ x }$

for the associated homogeneous

D = d/dx

$(D^2 + 1)y = 0$

roots a = b = i

Trial solution:

$y = (A +Bx)e^ix$

when you sub it into the homogeneous, the real parts cancel but the imaginary doesn't!

$-Ae^ix -Bxe^ix + 2Bie^ix +Ae^ix +Bxe^ix$

I was wondering why this is?

Am i right in thinking the trial solution is wrong?

For the particular solutions of the other parts I got

for $2cosx$

I got ${ y }_{ 2 } = x^2cosx$

for $2{ e }^{ x }$

I got ${ e }^{ x }$

thanks

2. ## Re: 2nd order differential complex!

The CF will be $Acosx+Bsinx,$ your trial solution should be $y=x(Ccosx+Dsinx)+Ee^{x}.$

3. ## Re: 2nd order differential complex!

When you have a complex conjugate pair (i and -i) of roots, then you use the cosine and sine functions. In your case, the solution to the homogeneous equation is $y=c_1cos{x}+c_2\sin{x}$.

Your particular solution for $2\cos{x}$ is incorrect. Try $x\sin{x}$.

- Hollywood