# help with variation of parameters question

• Apr 14th 2011, 11:09 PM
linalg123
help with variation of parameters question
Question: solve: 2x^2y'' + 5xy' + y = x^2 - x

My attempt so far: y'' + (5y'/2x) + (y/x^2) = 1/2 - 1/(2x)

yh: λ^2 + (5/2)λ +1 = 0
(2λ+1)(λ+2) = 0
λ = -1/2, -2

yh= c1e^(-1/2x) + c2e^(-2x)

Try yp = Ae^(-1/2x) + Bxe^(-1/2x) + C, but e^(-1/2x) is part of yh.

yp=
Axe^(-1/2x) + Bx^2e^(-1/2x) + C

is this correct so far?
• Apr 15th 2011, 04:08 AM
Jester
Answer - No! The method you're trying is the method of undetermined coefficients. In this method we "guess" the form of $y_p$. In your case since the RHS is quadratic try a general quadratic.
• Apr 15th 2011, 04:36 AM
topsquark
Quote:

Originally Posted by linalg123
Question: solve: 2x^2y'' + 5xy' + y = x^2 - x

My attempt so far: y'' + (5y'/2x) + (y/x^2) = 1/2 - 1/(2x)

yh: λ^2 + (5/2)λ +1 = 0
(2λ+1)(λ+2) = 0
λ = -1/2, -2

yh= c1e^(-1/2x) + c2e^(-2x)

Try yp = Ae^(-1/2x) + Bxe^(-1/2x) + C, but e^(-1/2x) is part of yh.

yp=
Axe^(-1/2x) + Bx^2e^(-1/2x) + C

is this correct so far?

Not to mention that the homogeneous solution method is incorrect. You can only use that for constant coefficients. Use the substitution $x = e^t$ to put it into constant coefficient format.

-Dan
• Apr 20th 2011, 04:32 PM
linalg123
i got the homogeneous solution is c1x^(-1/2) + c2x^(-1)

for the particular solution i tried
V1'x^(-1/2) + v2'x^(-1) = 0
-1/2v1'x^(-3/2) - v2'x^(-2) = x^2 - x

but i got the wrong answer.. is this the right approach?
• Apr 20th 2011, 04:57 PM
Prove It
Did you do what topsquark suggested? You can NOT find the solution to the homogenous solution the way you did, because it only works for constant coefficients.

You need to first make the substitution x = e^t --> t = ln(x).

So dy/dx = (dy/dt)(dt/dx) = (dy/dt)(1/x) = e^(-t)(dy/dt).

Get a similar expression for the second derivative, substitute and go from there.
• Apr 20th 2011, 05:27 PM
linalg123
so y'' = [(e^-t)(dy/dt) + (e^-t)(d^2y/dt^2)](e^-t)
y''= (e^-2t)(dy/dt) + (e^-2t)(d^2y/dt^2)

y' = e^(-t)(dy/dt)

y= ?

2e^(2t)[(e^-2t)(dy/dt) + (e^-2t)(d^2y/dt^2)] + 5e^t[e^(-t)(dy/dt)] + y = e^2t - e^t

2dy/dt + 2(d^2y/dt^2) + 5(dy/dt) + y = e^2t - e^t

how do i get rid of y?
• Apr 21st 2011, 06:58 AM
topsquark
Quote:

Originally Posted by linalg123
so y'' = [(e^-t)(dy/dt) + (e^-t)(d^2y/dt^2)](e^-t)
y''= (e^-2t)(dy/dt) + (e^-2t)(d^2y/dt^2)

Close. y'' = -e^{-2t}(dy/dt) + e^{-2t}(d^2y/dt^2)

I eventually get
2(d^2y/dt^2) + 3(dy/dt) + y = e^{2t} - e^t

Now you can solve the homogeneous equation. I get
yh(t) = Ae^{-t/2} + Be^{-t}

so your particular solution is going to be of the form
yp(t) = Ce^{2t} + De^t

-Dan