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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.

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Those are the coefficients of $e^{2t}$ in the original differential equation. – Sammy Black Apr 1 '13 at 6:33
By direct substitution into the left-hand side of the DE. – André Nicolas Apr 1 '13 at 6:35
up vote 3 down vote accepted

\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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Thank you. I feel ashamed that I did not see this. – 40Plot Apr 1 '13 at 6:38

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