1. ## Undetermined Coefficients

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.

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

Is the original equation satisfied?