Undetermined Coefficients

**Naples** · April 10th 2011, 09:48 PM

Solve the initial value problem using undetermined coefficients

$2y" + 3y' - 2y = 14x^2 - 4x - 11$ ; $y(0)=0 , y'(0)=(0)$
Use auxiliary to solve for $Yc = c1e^(^1^/^2^)^x + c2e^-^2^x$

Assume $Yp = Ax^2 + Bx + C$
$Yp' = 2Ax + B$
$Yp" = 2A$
Plug into original equation:
$4A + 6Ax + 3B - 2Ax^2 - 2Bx - 2C = 14x^2 - 4x - 11$
$-2Ax^2 = 14x^2$
$A = -7$

$6Ax - 2Bx = -4x$
$B = -19$

$4A + 3B - 2C = -11$
$C=-37$

So $Yp = -7x^2 - 19x - 37$

$Y = c1e^(^1^/^2^)^x + c2e^-^2^x - 7x^2 - 19x - 37$
$Y' = (1/2)c1e^(^1^/^2^)^x - 2c2e^-^2^x - 7x - 19$
$Y(0) = c1 + c2 - 37 = 0$
$c1 = 37 - c2$
$Y'(0) = (1/2)c1 - 2c2 - 19 = 0$

$(1/2)(37 - c2) - 2c2 = 19$
$c2 = -(1/5)$
$c1 = 186/5$

$Y = (186/5)e^(^1^/^2^)^x - (1/5)e^-^2^x - 7x^2 - 19x - 37$

I'd appreciate it if someone could check through my work and make sure I didn't make an error somewhere, thanks in advance.

**pickslides** · April 10th 2011, 09:52 PM

You can check it yourself by taking the derivative twice and substituting these results back into the original equation.

Is the original equation satisfied?

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