# Math Help - general solution of this differential equation?

1. ## general solution of this differential equation?

what's the general solution?

x*y*y' = y^2 + x*sqrt(4*x^2 + y^2)

2. Originally Posted by fryfry
what's the general solution?

x*y*y' = y^2 + x*sqrt(4*x^2 + y^2)
Note that if you divide both sides by x^2 the DE can be re-written as

$\displaystyle \frac{y}{x} \frac{dy}{dx} = \left( \frac{y}{x} \right)^2 + \sqrt{4 + \left( \frac{y}{x} \right)^2}$.

3. I'm not sure I quite agree. I get

$\displaystyle{\frac{d}{dx}\left(\frac{y^{2}}{x^{2} }\right)=\frac{2x^{2}yy'-2xy^{2}}{x^{4}}=\frac{2xyy'-2y^{2}}{x^{3}}=\frac{2}{x}\left(\frac{xyy'-y^{2}}{x^{2}}\right)=\frac{2}{x}\sqrt{4+\left(\fra c{y}{x}\right)^{2}}.}$

That last equality comes from the DE, from arranging the LHS to look like what's in the parentheses. At this point, you could make the substitution $u(x)=y(x)/x.$

I was thinking quotient rule for derivatives, but was getting hung up on not having the x squared term in the denominator, which you have to have in order to make the coefficients the same in the numerator.

4. Originally Posted by Ackbeet
I'm not sure I quite agree. I get

$\displaystyle{\frac{d}{dx}\left(\frac{y^{2}}{x^{2} }\right)=\frac{2x^{2}yy'-2xy^{2}}{x^{4}}=\frac{2xyy'-2y^{2}}{x^{3}}=\frac{2}{x}\left(\frac{xyy'-y^{2}}{x^{2}}\right)=\frac{2}{x}\sqrt{4+\left(\fra c{y}{x}\right)^{2}}.}$

That last equality comes from the DE, from arranging the LHS to look like what's in the parentheses. At this point, you could make the substitution $u(x)=y(x)/x.$

I was thinking quotient rule for derivatives, but was getting hung up on not having the x squared term in the denominator, which you have to have in order to make the coefficients the same in the numerator.
But I'm not differentiating the y^2/x^2, all I did was divide both sides by x^2. The DE in post #2 is correct (well, it is now that I put the = sign in), and the next step is to substitute y = xv.

5. Fair enough. I think the previous lack of equals sign was throwing me off.

6. $xyy'=y^2+x\sqrt{4*x^2 + y^2}$

May be this way is good
(1) $u^2=4x^2+y^2$
$uu'=4x+yy'$

The initial equation becomes
$u'-\frac{u}{x}=1$
$\frac{u}{x}=ln|x|+C$
From (1) get y.