Question

Solutions for y'= x²+y²/2xy

Answer 1

Let us solve the differential equation $y'= \frac{x²+y²}{2xy}$.

Let $y=ux$. Then $y'=u'x+u$. Consequently we have the following differential equation:

$u'x+u=\frac{x²+u²x^2}{2ux^2}=\frac{1+u²}{2u}$

$u'x=\frac{1+u²}{2u}-u=\frac{1-u²}{2u}$

$\frac{du}{dx}x=\frac{1-u²}{2u}$

$\frac{2udu}{1-u^2}=\frac{dx}{x}$

$\int\frac{2udu}{1-u^2}=\int\frac{dx}{x}$

$-\int\frac{d(1-u^2)}{1-u^2}=\int\frac{dx}{x}$

$-\ln|1-u^2|=\ln|x|-\ln|C|$

$\ln|C|=\ln|x(1-u^2)|$

$C=x(1-u^2)$

$C=x(1-\frac{y^2}{x^2})$

Therefore, the solution of the differential equation $y'= \frac{x²+y²}{2xy}$ is the following:

$C=x-\frac{y^2}{x}$