Finding Particular Solution of a Second Order Differential equation

Question

The homogenous equation: $\ 28{e^{-2x}} { - 18e^{-3x}}$

I found the homogenous solution to the equation, however I am not sure how to find the particular solution when the differential equation is equal to 8. I tried using matlab but I'm not too sure if it's correct.

Answer 1

Because $v_s$ is a constant we have $f(v'',v',v)=P(t)$ where $P$ is polynomial with degree $n$ (and $f$ is linear) . In this particular case $P$ is degree $0$. A second order ODE in this form has praticular solution in the form of $Q(x)$, where $Q$ is polynomial in the same degree as $P$, so in this case degree $0$.

Hence $Q(x)=a$:

$(a)''+5(a)'+6a=8\implies 6a=8\implies a=4/3$ so $v_p=4/3$

Answer 2

If $v_s$ is constant, make the substitution $v = u + v_s/6.$ Then $u$ satisfies the homogeneous equation and you are done. If $v_s$ is not necessarily constant, use variation of parameters: look for solution in the form $v = v_0 f,$ where $v_0$ is the homogeneous solution.