if i have a problem...

$\displaystyle

y'' + 2y' + 4 = x^2 e^{ - 2x}

$

would the substitution be..

$\displaystyle

y(x) = e^{ - 2x} v

$

I know it would be for = xe^-2x..

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- May 22nd 2009, 06:34 AMdankelly07just a quick particular integral question..
if i have a problem...

$\displaystyle

y'' + 2y' + 4 = x^2 e^{ - 2x}

$

would the substitution be..

$\displaystyle

y(x) = e^{ - 2x} v

$

I know it would be for = xe^-2x.. - May 22nd 2009, 06:49 AMTheEmptySet
- May 22nd 2009, 07:11 AMHallsofIvy
It's not clear what you are asking. You seem to be conflating "variation of parameters" and "undetermined coefficients".

Assuming you mean $\displaystyle y"+ 2y'+ 4y= x^2e^{-2x}$, which has $\displaystyle e^{-2x}$ and $\displaystyle xe^{-2x}$ as independent solutions to the associated homogenous equation, in order to use "variation of parameters" you would have to use $\displaystyle y(x)= u(x)e^{-x}+ v(x)x^2e^{-x}$. If you are referring to "undetermined coefficients", you would try $\displaystyle y(x)= (Ax^2+ bx+ C)x^2e^{-x}= (Ax^4+ Bx^3+ Cx^2)e^{-x}$.