Homework Help: Finding particular solution yp of given equation

1. Sep 6, 2011

ProPatto16

y''+2y'-3y=1+xex

ive tried yp=Axex+B

and yp=Ax2ex+Bx

and both dont work. assuming im doing it correctly.

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

suggestions?

thanks

2. Sep 6, 2011

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.

3. Sep 6, 2011

LCKurtz

The proper choice is to try yp = Axex+Bx2ex.

4. Sep 6, 2011

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.

5. Sep 7, 2011

ProPatto16

heyy...

i tried yp=Axex+Bx2ex and got

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

result was A(4ex)+B(8xex+2ex)=1+xex

looking at that i saw i could cancel out constant A by making it Aex

using Aex+Bxxex

after subbing in i got B(8xex+2ex)=1+xex

how can i possibly solve for A or B??

6. Sep 7, 2011

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.

7. Sep 7, 2011

ProPatto16

i dont 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?

8. Sep 7, 2011

HallsofIvy

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

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

9. Sep 7, 2011

ProPatto16

ohhh. I missed that :/

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

10. Sep 7, 2011

ProPatto16

y= Axex+bx2ex

y'= A(xex+ex) + B(2xex+x2ex)

y''= A(xex+ex+ex) + B(2xex+2ex+2xex+x2ex)

subbing in

y''+2y'-3y = [A(xex+ex+ex) + B(2xex+2ex+2xex+x2ex)] + 2[A(xex+ex) + B(2xex+x2ex)] - 3[Axex+bx2ex]

= A[xex+2ex+2xex+2ex-3xex] + B[4xex+2ex+x2ex+4xex+2x2ex-3x2ex

= A[4ex] + B[8xex+2ex]

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

it seems like one of my 2xex should only be 2ex so i get 6 and 4 instead of 8 and 2....

11. Sep 7, 2011

HallsofIvy

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

12. Sep 7, 2011

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! :)

13. Sep 7, 2011

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

14. Sep 7, 2011

ProPatto16

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

y''+2y'-3y=1+xex

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