Use the method of undetermined coefficients to solve the following differential equation: $\displaystyle y''+y'=4x$
$\displaystyle y(x)=$______$\displaystyle +C1+$______$\displaystyle C2$_____
I'm a little rusty on Undetermined Coefficients, but you want to start by arbitrarily choosing a function in the same form as the inhomogeneous one you're giving.
For example, $\displaystyle y_{p}=ax+b$ should work for this situation.
Find the derivative, and second derivative so that
$\displaystyle y_{p}'=a$
and $\displaystyle y_{p}''=0$
Now that you have that, you can plug in these functions where they are needed in the original. So, you should get
$\displaystyle a+0=4x$
I'm sort of confused, and maybe you left out a y factor, but if this is right, then you get 0 for both coefficients.
So, assuming that you did write the original equation correctly, then the solution to the inhomogeneous part is as follows: $\displaystyle y_{p}(x)=0$
You can solve for $\displaystyle y_{h}(x)$ using your typical method of solving for $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$
If you need help with the homogeneous solution, just post, and I will help you out.