1. ## Undetermined Coefficients

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.

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?