# Thread: Second order linear ODE

1. ## Second order linear ODE

Just wondering how would you find the particular solution to the following differential equation. Is there a quick technique or way of doing this?

$2y''-3y'-5y=(x+4)e^{5x/2}$

2. Inspired guesswork.

Note that the right hand side can be expanded as $x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$.

So an appropriate particular solution could be $y_p = Ax\,e^{\frac{5x}{2}} + Be^{\frac{5x}{2}}$.

This means $y_p' = Ae^{\frac{5x}{2}} + \frac{5}{2}Ax\,e^{\frac{5x}{2}} + \frac{5}{2}Be^{\frac{5x}{2}}$

$= \left(A + \frac{5}{2}B\right)e^{\frac{5x}{2}} + \frac{5}{2}Ax\,e^{\frac{5x}{2}}$

and $y_p'' = \frac{5}{2}\left(A + \frac{5}{2}B\right)e^{\frac{5x}{2}} + \frac{5}{2}Ae^{\frac{5x}{2}} + \frac{25}{4}Ax\,e^{\frac{5x}{2}}$

$= \left(5A + \frac{25}{4}B\right)e^{\frac{5x}{2}} + \frac{25}{4}Ax\,e^{\frac{5x}{2}}$.

Substituting into the DE gives...

$2\left[\left(5A + \frac{25}{4}B\right)e^{\frac{5x}{2}} + \frac{25}{4}Ax\,e^{\frac{5x}{2}}\right] - 3\left[\left(A + \frac{5}{2}B\right)e^{\frac{5x}{2}} + \frac{5}{2}Ax\,e^{\frac{5x}{2}}\right] - 5\left(Ax\,e^{\frac{5x}{2}} + Be^{\frac{5x}{2}}\right) = x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$

$\left(10A + \frac{25}{2}B\right)e^{\frac{5x}{2}} + \frac{25}{2}Ax\,e^{\frac{5x}{2}} + \left(-3A - \frac{15}{2}B\right)e^{\frac{5x}{2}} - \frac{15}{2}Ax\,e^{\frac{5x}{2}} - 5Ax\,e^{\frac{5x}{2}} - 5Be^{\frac{5x}{2}} = x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$

$5Ax\,e^{\frac{5x}{2}} + \left(7A + 5B\right)e^{\frac{5x}{2}} = x\,e^{\frac{5x}{2}} + 4e^{\frac{5x}{2}}$.

Therefore $5A = 1$ and $7A + 5B = 4$.

This gives $A = \frac{1}{5}$

and $\frac{7}{5} + 5B = 4$

$5B = \frac{13}{5}$

$B = \frac{13}{25}$.

So finally, we have our particular solution

$y_p = \frac{1}{5}x\,e^{\frac{5x}{2}} + \frac{13}{25}e^{\frac{5x}{2}}$.