What is the solution for y^'= x/(3y) ?

The solution for y^'= x/(3y) is y=sqrt((x^2)/3+c_{1)},y=-sqrt((x^2)/3+c_{1)}

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y^'= x/(3y)

y^{^{'}} = \frac{x}{3 y}

y = \frac{x ^{2}}{3} + c_{1}, y = - \frac{x ^{2}}{3} + c_{1}

Solution steps

y^{'} (x) = \frac{x}{3 y}

Solve separable ODE:

y = \frac{x ^{2}}{3} + c_{1}, y = - \frac{x ^{2}}{3} + c_{1}

y = \frac{x ^{2}}{3} + c_{1}, y = - \frac{x ^{2}}{3} + c_{1}

\int \frac{6}{1 - cos ( 3 x )} d x

\frac{( x ^{2} + 4 x + 3 )}{x}

\int x 1 0^{x} d x

(\frac{9 y ^{2} - t ^{2}}{y ^{5}}) \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

x^{2} - 2 x + 4, 2

What is the solution for y^'= x/(3y) ?
The solution for y^'= x/(3y) is y=sqrt((x^2)/3+c_{1)},y=-sqrt((x^2)/3+c_{1)}