I have an IVP I'm trying to solve with Laplace transformations:
$$y''-4y'+4y=te^{2t}$$
Given that: $y(0)=1$ and $y'(0)=0$
I've gotten to the part where I isolate $Y(s)$:
(1)$$Y(s)=\frac{(s-4)}{(s-2)^2}+\frac{1}{(s-2)^4}$$
(2)
$$Y(s)=\frac{(s-4)(s-2)^2+1}{(s-2)^4}$$
Now, obviously it is time for partial fraction expansion:
$$Y(s)=(s-4)(s-2)^2+1=A(s-2)^3+B(s-2)^2+C(s-2)+D$$
Immediately, just by substituting $s=2$, we can get 1 for the value of D. However we can't do s-substitution to isolate A, B or C. So we have to equate the coefficients.
However, expanding all this out given that there are things like $A(s-2)^3$, would be really, really tedious so I'm wondering if there is some shortcut to this.
This question was given on a test years ago and the professor who wrote the answer and working out magically skipped from (1) to all the partial fractions having A, B, C & D already solved. It was written like so:
$$Y(s)=\frac{(s-4)}{(s-2)^2}+\frac{1}{(s-2)^4}=\frac{1}{(s-2)}-\frac{2}{(s-2)^2}+\frac{1}{(s-2)^4}$$
Part of me thinks the professor simply skipped writing out all the tedious equating of the coefficients for the sake of lazyness however given that the test is only an hour long, this question takes up too much time so I'm wondering if there is some glaringly obvious shortcut he used that I'm missing.
Don't worry about the answer or looking up the tables of the transforms, I've already done this. I just need to know if there's a super secret rule of thumb to skip doing 10 years of equating of the coefficients to solve A, B and C.
EDIT: As Przemysław Scherwentke hinted, you can "avoid" the full partial fraction expansion of this question in this manner:
$$Y(s)=\frac{(s-4)}{(s-2)^2}+\frac{1}{(s-2)^4}=\frac{(s-2)-2}{(s-2)^2}+\frac{1}{(s-2)^4}$$
$$Y(s)=\frac{(s-2)}{(s-2)^2}-\frac{2}{(s-2)^2}+\frac{1}{(s-2)^4}$$
The first fraction reduces…
$$Y(s)=\frac{1}{(s-2)}-\frac{2}{(s-2)^2}+\frac{1}{(s-2)^4}$$
Thus there is no need for partial fraction decomposition as it is already of the form:
$$\frac{A}{(s-2)}+\frac{B}{(s-2)^2}+\frac{C}{(s-2)^3}+\frac{D}{(s-2)^4}$$
Where $A = 1, B = -2, C = 0$ since the fraction with the denominator $(s-2)^3$ is absent and $D = 1$
Best Answer
HINT: $s-4=(s-2)-2$ and you can reduce something... But generally expansion is not obvious, hence your undesired calculations.