[Math] Particular Solution of $y” – 3y’ – 4y = 3e^{2t}$

ordinary differential equations

I took this example out of the textbook but I am unable to understand one part – after hours staring at it.

For just the particular solution,
$$ Y(t) = Ae^{2t} $$
$$ Y'(t) = 2Ae^{2t} $$
$$ Y''(t)=4Ae^{2t} $$

Here is the part where I don't quite understand:
$$ (4A-6A-4A)e^{2t} = 3e^{2t}$$

I know we are equating the RHS of the differential equation with the particular solution to solve for A, but I'm not sure how did the textbook get $(4A-6A-4A)$ from.

Best Answer

\begin{align*} y'' - 3y' - 4y &= 4A e^{2t} - 3\cdot 2Ae^{2t} - 4\cdot Ae^{2t} &\text{by the equalities above} \\ &= (4A - 6A - 4A)e^{2t} &\text{by multiplication} \\ &= -6Ae^{2t} &\text{simplification} \\ y'' - 3y' - 4y &= 3e^{2t} &\text{the differential equation} \end{align*}

It should be clear that the two sides are equal from the reasoning above. For the first equality, we're just substituting the equalities that have been given.

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