1. ## [SOLVED] homogeneous DEQ

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

let
$\displaystyle y = vx$

$\displaystyle dy = vdx + xdv$

and after that i substitute and got a result of

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

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

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

how do i delete this thread? thanks

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

let
$\displaystyle y = vx$

$\displaystyle dy = vdx + xdv$

and after that i substitute and got a result of

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

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

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

3. $\displaystyle x^2y' = (4x^2 + 7xy + 2y^2)$

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

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

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

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

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

integrate the left side by partial fraction

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

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

substitute y = vx

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

$\displaystyle y + x = cx^2(y+2x)$

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

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

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

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

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

-Dan