A few days ago
find the particular solution of the differential equation: y” – 4y’ + 4y = e^(2x)?
I’m not sure what to do b/c I get 0 for A?? Can someone help, this is what I had so far:
Y(x) = Ae^2x
Y'(x) = 2Ae^2x
Y”(x) = 4Ae^2x
So… 4Ae^2x – 8Ae^2x + 4Ae^2x = e^2x
Which becomes 0 = e^2x
and then I’m stumped…
Top 1 Answers
A few days ago
Favorite Answer
Since your A becomes a 0, you can “guess” your particular solution to be y(x)= A*x*e^2x. That is, you just multiple your initial solution with another x.
Then, y'(x)= Ae^2x+2Axe^2x and so forth.
Good luck!
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