Find a Particular Solution 1 Xe X

You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.

Forums
Homework Help
Calculus and Beyond Homework Help

Finding particular solution yp of given equation

Thread starter ProPatto16
Start date Sep 6, 2011

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

ive tried y_p=Axe^x+B

and y_p=Ax²e^x+Bx

and both dont work. assuming im doing it correctly.

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

suggestions?

thanks

Answers and Replies

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

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

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.

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??

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

Solve for A and B.

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?

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

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

ohhh. I missed that :/

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

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....

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

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! :)

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

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 ??

Hentze Youche1973

Find a Particular Solution 1 Xe X

Finding particular solution yp of given equation

Answers and Replies

Related Threads on Finding particular solution yp of given equation

Menu Halaman Statis