# [SOLVED] homogeneous DEQ

• Nov 28th 2007, 07:42 AM
[SOLVED] homogeneous DEQ
obtain the family of solutions
$x^2y' = 4x^2 + 7xy + 2y^2$

let
$y = vx$

$dy = vdx + xdv$

and after that i substitute and got a result of

$xdv + vdx = (4 + 7v + v^2)dx$

$\frac{dv}{4 + 6v + v^2}= \frac{dx}{x}$
i dont know if what i did is right .... im stuck with integrating the left part of this equation
thank you

edit
$xdv + vdx = (4 + 7v + 2v^2)dx$
i think i already know what's wrong with my soln

how do i delete this thread? thanks
• Nov 29th 2007, 05:30 AM
topsquark
Quote:

obtain the family of solutions
$x^2y' = 4x^2 + 7xy + 2y^2$

let
$y = vx$

$dy = vdx + xdv$

and after that i substitute and got a result of

$xdv + vdx = (4 + 7v + v^2)dx$

$\frac{dv}{4 + 6v + v^2}= \frac{dx}{x}$
i dont know if what i did is right .... im stuck with integrating the left part of this equation
thank you

edit
$xdv + vdx = (4 + 7v + 2v^2)dx$
i think i already know what's wrong with my soln

how do i delete this thread? thanks

Rather than delete it, why don't you tell us how you solved it? Your answer might be helpful to others.

-Dan
• Nov 29th 2007, 05:48 AM
$x^2y' = (4x^2 + 7xy + 2y^2)$

let $y = vx$
$dy = vdx + xdv$

$x^2dy = (4x^2 + 7xy + 2y^2)dx$

$x^2(xdv + vdx) = (4x^2 + 7x^2v + 2x^2v^2)dx$

$(xdv + vdx) = (4 + 7v + 2v^2)dx$

$\frac{dv}{4 + 6v + 2v^2} = \frac{dx}{x}$

integrate the left side by partial fraction

$\frac{1}{2}(-\ln|v+2| + \ln|v+1|) = \ln{x} + \ln{c}$
$-ln|v+2| + \ln|v+1| = 2lnx + c$

$\frac{v+1}{v+2} = x^2c$

substitute y = vx

$\frac{y}{x} + 1 = cx^2(\frac{y}{x} + 2)$

$y + x = cx^2(y+2x)$
• Nov 29th 2007, 07:23 AM
topsquark
Quote:

$\frac{1}{2}(-\ln|v+2| + \ln|v+1|) = \ln{x} + \ln{c}$
$-ln|v+2| + \ln|v+1| = 2lnx + c$

$\frac{v+1}{v+2} = x^2c$

The only thing I have to say about this, and I would imagine it's likely due to a typo, is your treatment of the arbitrary constant in these three lines. I would write something like
$\frac{1}{2}(-\ln|v+2| + \ln|v+1|) = \ln{x} + \ln{c}$

$-ln|v+2| + \ln|v+1| = 2lnx + ln(c^{\prime})$ <-- Where $c^{\prime}$ is a new constant.

$\frac{v+1}{v+2} = x^2c^{\prime}$

or just keep the original constant and wind up with
$\frac{v+1}{v+2} = x^2c^2$
(This last form might even be seen to be the best because it illustrates that the constant $c^{\prime}$ is required to be positive.)

-Dan