Solve the initial value problem using undetermined coefficients

$\displaystyle 2y" + 3y' - 2y = 14x^2 - 4x - 11$ ; $\displaystyle y(0)=0 , y'(0)=(0)$

Use auxiliary to solve for $\displaystyle Yc = c1e^(^1^/^2^)^x + c2e^-^2^x$

Assume $\displaystyle Yp = Ax^2 + Bx + C$

$\displaystyle Yp' = 2Ax + B$

$\displaystyle Yp" = 2A$

Plug into original equation:

$\displaystyle 4A + 6Ax + 3B - 2Ax^2 - 2Bx - 2C = 14x^2 - 4x - 11$

$\displaystyle -2Ax^2 = 14x^2$

$\displaystyle A = -7$

$\displaystyle 6Ax - 2Bx = -4x$

$\displaystyle B = -19$

$\displaystyle 4A + 3B - 2C = -11$

$\displaystyle C=-37$

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

$\displaystyle Y = c1e^(^1^/^2^)^x + c2e^-^2^x - 7x^2 - 19x - 37$

$\displaystyle Y' = (1/2)c1e^(^1^/^2^)^x - 2c2e^-^2^x - 7x - 19$

$\displaystyle Y(0) = c1 + c2 - 37 = 0$

$\displaystyle c1 = 37 - c2$

$\displaystyle Y'(0) = (1/2)c1 - 2c2 - 19 = 0$

$\displaystyle (1/2)(37 - c2) - 2c2 = 19$

$\displaystyle c2 = -(1/5)$

$\displaystyle c1 = 186/5$

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

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