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

Dots

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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find the particular solution of the differential equation: y” – 4y’ + 4y = e^(2x)?

find the particular solution of the differential equation: y” – 4y’ + 4y = e^(2x)?

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