Coefficients of homogeneous second order ODE when initial conditions are both 0

Question

I'm currently trying to solve this problem: Find the unique solution to the differential equation $y''- y'- 6y = 18x$ With initial conditions $y'(0) = 0 , y(0) = 0$

My current solution is $y = -3x + 1/2 + Ae^{3x} + Be^{-2x}$ and I'm struggling at the last step of finding $A$ and $B$.

Any help would be greatly appreciated

Answer 1

Hint: Plug in $x=0$ into your final equation $y(x=0)=...=0$. And then plug in $x=0$ into the derivative of $y$. You will have the second equation $y'(x=0)=...=0$. Now solve this system of linear equations for $A$ and $B$.

Answer 2

The particular solution to $ y''(x) - 6y(x) - y' = 18x $ is of the form :

$ y_p(x) = a_1 + a_2x $

$ y_p'(x) = a_2 $ and $ y_p''(x) = 0 $

Substitute the particular solution $y_p(x)$ into the differential eq. :

$-a_2 - 6(a_1 + a_2x) = 18x $

Equate the coefficients of 1 on both sides of the equation :

$-6a_2 = 18 \Leftrightarrow a_2 = -3 $ and you will also have $a_1=\frac{1}{2}$

Substitute a_1 and a_2 into $ y_p(x) = a_1 + a_2x $

You get : $ y_p(x) = \frac{1}{2} -3 x $

So your solution is : $ y(x) = -3x + c_1e^{-2x} + c_2e^{3x} + \frac{1}{2}$