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

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

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y^'=(3x)/y

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (\frac{ln ( 3 x ^{2} - 6 x )}{4 x ^{3} - 2})

\int (ln (x))^{7} d x

\frac{d}{d x} (4 sec (4 x))

\int e^{(x + 1)} d x

n = 1 \sum \infty cos (\frac{1}{n})

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