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Upon reaching the point in Differential Equations where I am attempting to solve a non-homogeneous ODE to reach a general solution, I have become quite confused. For sake of example, I have the ODE: $$y''+3y'+2y = 3e^{-4t}$$ $y(0)=1 , y'(0)=0$

CASE 1: I believe I learned that it is possible to solve the left side of the ODE as if it was a homogeneous equation, then solve for the particular solution of the ODE. The general solution would then be a sum of the particular solution and homogeneous equation.

CASE 2: However, all the answers to such types of problems have found the particular solution first, then found the general solution by using the particular solution as part of finding the homogeneous solution.

Would CASE 1 be correct as well?

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The point is that you need to find some (any) $Y(t)$ that satisfies the whole equation $y''+3y'+2y = 3e^{-4t}$. Then if you find a $Z(t)$ that satisfies $y''+3y'+2y = 0$ you have that $Y(t)+Z(t)$ will also satisfy your equation. Plug and chug to verify that. This works because your equation is linear. You find a 2D linear space of solutions to the homogeneous equation.

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You should do case 1 either way, to find general solution. Case 2 enables you to find the correct constants for a particular solution (for a given y(0),y'(0)) so you substitute them in the solution after case 1. Hope that helps..

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