Finding particular solution yp of given equation

[ProPatto16]

y''+2y'-3y=1+xe^x

ive tried y_p=Axe^x+B

and y_p=Ax²e^x+Bx

and both don't work. assuming I am doing it correctly.

im given the answer as: 1/3+1/16(2x²-x)e^x

suggestions?

thanks

 

[HallsofIvy]

$e^x$ is a solution to the associated homogeneous equation so if the function on the right were just $e^x$ you would have to try $xe^x$. Because you want $xe^x$, try $Ax^2e^x$ instead.

 

[LCKurtz]

The proper choice is to try y_p = Axe^x+Bx²e^x.

 

[ProPatto16]

Thanks guys, LC that makes sense now I see it, I was varying the order of the ex^x but not the 1. Which of course is silly.

 

[ProPatto16]

heyy...

i tried y_p=Axe^x+Bx²e^x and got

after subbing back into y''+2y'-3y

result was A(4e^x)+B(8xe^x+2e^x)=1+xe^x

looking at that i saw i could cancel out constant A by making it Ae^x

using Ae^x+Bx^xe^x

after subbing in i got B(8xe^x+2e^x)=1+xe^x


how can i possibly solve for A or B??

 

[HallsofIvy]

$y= (Ax+ Bx^2)e^x$
$y'= (A+ 2Bx)e^x+ (Ax+ Bx^2)e^x= (A+ (2B+ A)x+ Bx^2)e^x$
$y''= (2B+ A+ 2B)e^x+ (A+ (2B+ A)x+ Bx^2)e^x= (6B+ 2A+ (2B+ A)x+ Bx^2)e^x$
$y''+ 2y'- 3y= [(6B+ 4A)+ 4Bx]e^x= xe^x$

Solve for A and B.

 

[ProPatto16]

i don't know how to do the quotes thing so ill copy and paste.

" y′′+2y′−3y=[(6B+4A)+4Bx]ex=xex "


but from the original question... y′′+2y′−3y=xex +1

whered the +1 go?

 

[HallsofIvy]

Because the equation is linear, you can do that separately.

Let y= B. Then y''= y'= 0 so the equation becomes ...

 

[ProPatto16]

ohhh. I missed that :/

hang around please, gimme 10min to post the way i did it. its slightly different.

 

[ProPatto16]

y= Axe^x+bx²e^x

y'= A(xe^x+e^x) + B(2xe^x+x²e^x)

y''= A(xe^x+e^x+e^x) + B(2xe^x+2e^x+2xe^x+x²e^x)

subbing in

y''+2y'-3y = [A(xe^x+e^x+e^x) + B(2xe^x+2e^x+2xe^x+x²e^x)] + 2[A(xe^x+e^x) + B(2xe^x+x²e^x)] - 3[Axe^x+bx²e^x]

= A[xe^x+2e^x+2xe^x+2e^x-3xe^x] + B[4xe^x+2e^x+x²e^x+4xe^x+2x²e^x-3x²e^x

= A[4e^x] + B[8xe^x+2e^x]

the only thing that differs from your solution is the B parts

it seems like one of my 2xe^x should only be 2e^x so i get 6 and 4 instead of 8 and 2...

 

[HallsofIvy]

I had before, $y′′+2y′−3y=[(6B+4A)+4Bx]e^x=xe^x$
so you must have 4B= 1 and 6B+ 4A= 0.

 

[ProPatto16]

yeah i did that part. there just an error in my jumble up there. was hoping you'd see it. but nevermind. thanks for all your help though, much appreciated! :)

 

[ProPatto16]

yeah... i was right...

to get the correct answer the euqations for A and B should be

8B=1 and 2B+4A=0

gives B = 1/8 and A = -1/16

 

[ProPatto16]

about that +1 that got left alone... you said we can do that seperately...

y''+2y'-3y=1+xe^x

taking out all differentiable functions leaves 3y=-1 so y= -1/3 + y_p ??