What is the solution for y^'= x/y+y/x ?

The solution for y^'= x/y+y/x is y=xsqrt(2ln(x)+c_{1)},y=-xsqrt(2ln(x)+c_{1)}

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y^'= x/y+y/x

y^{^{'}} = \frac{x}{y} + \frac{y}{x}

y = x 2 ln (x) + c_{1}, y = - x 2 ln (x) + c_{1}

Solution steps

y^{'} (x) = \frac{x}{y} + \frac{y}{x}

Solve Bernoulli ODE:

y = x 2 ln (x) + c_{1}, y = - x 2 ln (x) + c_{1}

y = x 2 ln (x) + c_{1}, y = - x 2 ln (x) + c_{1}

\int 3 sec^{2} (3 x) d x

\frac{d}{d x} ((x^{2})^{3})

\frac{d}{d x} (\frac{sec ( x )}{1 + tan ( x )})

\frac{d}{d x} (- 0.0006 x (x - 15)^{2} (x - 40))

d er i v a t i v e y = x (7 x + 3)^{\frac{1}{3}}

What is the solution for y^'= x/y+y/x ?
The solution for y^'= x/y+y/x is y=xsqrt(2ln(x)+c_{1)},y=-xsqrt(2ln(x)+c_{1)}