How to deduce constants from second order ODE that is derived from 2 linear ODE?

Question

These two linear ODEs: $$ \frac{du}{dx} + 2u= e^x $$ and $$ \frac{dv}{dx} -3v=2u $$ could be reduced to a single second order ODE for $v(x)$ of the form $$ \frac{d^2v}{dx^2} + a\frac{dv}{dx} + bv =Q(x). $$ Without doing any differentiations, deduce from your solution to $v$, the values of $a$ and $b$.

My work. I got $$ v(x)=Be^{3x} -( 2/5)Ae^{-2x} -(1/3)e^x $$ and $$ u(x)=Ae^{-2x} + (1/3)e^x.$$ But I really have no idea how to deduce constants $a$ and $b$? Please help.

Answer 1

$$ \left(\frac{d}{dx}+2\right)u=e^x\\ \left(\frac{d}{dx}-3\right)v=u $$

hence

$$ \left(\frac{d}{dx}+2\right)\left(\frac{d}{dx}-3\right)v = \left(\frac{d}{dx}+2\right)u=e^x $$

then

$$ \left(\frac{d^2}{dx^2}-\frac{d}{dx}-6\right)v = e^x $$

Answer 2

Hint. Consider the characteristic polynomials of the two given first order linear ODEs: $$(D+2I)u=e^x\;,\;(D-3I)v=2u\implies (D+2I)2u=(D+2I)(D-3I)v=(D-3I)2e^x.$$ What is the characteristic polynomial of the single second order linear ODE on the right?