Solve:
$$\huge y''+4y'+4y=6xe^{-2x} $$

Question

Solve:
$$\huge y''+4y'+4y=6xe^{-2x} $$

anonymous · Answer

Not again!

anonymous · Answer

Have you considered posting it w/o LaTeX?

anonymous · Answer

the roots are -2 so Yh = c1e^(-2x)+c2e^(-2x)
why multiply one term of the complimentary solution with an x such that it becomes
Yh = c1e^(-2x)x+c2e^(-2x)?

anonymous · Answer

That's the only thing I'm having trouble with

turingtest · Answer

to keep the solutions linearly independent

anonymous · Answer

So the Wronskian has to be non zero in order for solutions to exist?

turingtest · Answer

The Wronskian has to be non-zero in order for the solutions to form a fundamental set, i.e. to span the solution space of the problem

turingtest · Answer

do you still need help with this problem?
I would do variation of parameters...

amistre64 · Answer

6x is linear so constant coeffs is fine with a suitable fillin

anonymous · Answer

I used Yp = (A1x^3+A0x^2)e^(-2x)
I get A0=0 and A1=1

amistre64 · Answer

Ax e^-2x + Be^-2x

but we gots to go higher than an Ax

Ax^3 e^-2x + Bx^2e^-2x

amistre64 · Answer

yeah, that one

turingtest · Answer

I think your values for the constants are wrong cyter