d²y/dx² -4dy/dx + 4y=3e^2x

Question

Ordinary differential equations

Answer 1

This can be written: y"-4y'+4y=3e^2x.

First find the characteristic equation by solving r²-4r+4=0=(r-2)².

Since there's only one root r=2 then the solution is y_c=Ae^2x+xBe^2x, where y_c denotes the characteristic solution. Let's check this out:

y_c'=2Ae^2x+Be^2x+2xBe^2x; y_c"=4Ae^2x+4Be^2x+4xBe^2x; 

y_c"-4yc'+4y_c=

[4Ae^2x+4Be^2x+4xBe^2x]-4[2Ae^2x+Be^2x+2xBe^2x]+4[Ae^2x+xBe^2x]=

4Ae^2x+4Be^2x+4xBe^2x-8Ae^2x-4Be^2x-8xBe^2x+4Ae^2x+4xBe^2x=

4Ae^2x+4Ae^2x-8Ae^2x+4Be^2x-4Be^2x+4xBe^2x+4xBe^2x-8xBe^2x=0✔️

Now we need the particular solution y_p. Note that 3e^2x is another e^2x term. We've already used e^2x and xe^2x, so it would futile to use either of these.

However, ax²e^2x (a is another constant which can be calculated) hasn't been used and, furthermore, the second derivative of ax²e^2x would add the extraneous term 2ae^2x and therefore would not cancel out the way that e^2x and xe^2x terms did when deriving y_c.

Therefore 2ae^2x=3e^2x, making a=3/2.

y=y_c+y_p=Ae^2x+xBe^2x+(3/2)x²e^2x. This can be written y=(A+Bx+(3/2)x²)e^2x.