How to Solve Using Undetermined Coefficients: y+3y=-48x^2e^(3x)

[vipertongn]

Homework Statement

solve using undetermined coeffecients
y"+3y=-48x^2e^(3x)

Homework Equations

x^2e^3x gives a Yp=(Ax^2+Bx+C)e^3x

The Attempt at a Solution

I can get yc easily which is yc=cos(sqrt(3x))+sin(sqrt(3x))

However, I'm not sure as to how to set it up so that -48x^2e^(3x) and the yp. the -48 is throwing me off. If I make it so then that is equivalent to a constant D in teh equation I get

Yp= De^3x(Ax^2+Bx+C)
taking to the second derivative
Yp"+3Yp=3D[(10Ax^2e^2x)+(18Axe^3+7Bxe^3x)+(6Ae^3x+9Be^3x+6Ce^3x)=-48x^2e^(3x)
At most I could get D=-16

 

[HallsofIvy]

Homework Statement

solve using undetermined coeffecients
y"+3y=-48x^2e^(3x)

Homework Equations

x^2e^3x gives a Yp=(Ax^2+Bx+C)e^3x

The Attempt at a Solution

I can get yc easily which is yc=cos(sqrt(3x))+sin(sqrt(3x))

Be careful of your parentheses- and don't forget the unknown constants. yc= Acos(sqrt(3)x)+ B sin(sqrt(3)x) for any constants A and B.

However, I'm not sure as to how to set it up so that -48x^2e^(3x) and the yp. the -48 is throwing me off. If I make it so then that is equivalent to a constant D in teh equation I get

Yp= De^3x(Ax^2+Bx+C)
taking to the second derivative
Yp"+3Yp=3D[(10Ax^2e^2x)+(18Axe^3+7Bxe^3x)+(6Ae^3x+9Be^3x+6Ce^3x)=-48x^2e^(3x)
At most I could get D=-16

Okay, your general form for Yp is correct- any time you have a power of x on the right, you must try all lower powers as well.

But either you have some very strange ideas about the derivatives of exponentials or you have a lot of typos! You cannot possibly get e^(2x) or e^3 by differentiating e^(3x)!
And, in fact, "D", the one thing you say you can find, is unnecessary: you do not need a separate constant "D" multiplying the entire thing- you can take that inside the parentheses: Yp= e^(3x)(ADx^2+ BDx+ CD) or just Yp= e^(3x)(Ax^2+ Bx+ C) where the new A, B, C are the previous AD, BD, BC.

The first derivative of Yp is Yp'= 3e^(3x)(Ax^2+ Bx+ C)+ e^(3x)(2Ax+ B)= e^(3x)(3Ax^2+ (3B+ 2A)x+ 3C+ B).

The second derivative is Yp"= 3e^(3x)(3Ax^2+ (3B+ 2A)x+ 3C+ B)+ e^(3x)(6Ax+ 3B+ 2A)= e^(3x)(9Ax^2+ (12A+ 9B)x+ 2A+ 3B+ 9C)

Yp"+ 3Y= e^(3x)(10Ax^2+ (12A+ 10B)x+ 2A+ 3B+ 10C)= -48x^2e^(3x)= e^(3x)(-48x^2+ 0x+ 0).

You can cancel the "e^(3x)" on each side to reduce to 10Ax^2+ (12A+ 10B)x+ 2A+ 3B+ 10C= -48x^2+ 0x+ 0. In order to be equal for all x, two polynomials must have exactly the same coefficients:

10A= -48, 12A+ 10B= 0, and 2A+ 3B+ 10C= 0.

 

[Tanzy]

Hello! Sorry for the old thread digging but can you tell me from which book this problem is found?