What is the solution for y^'=(4x)/y ?

The solution for y^'=(4x)/y is y=sqrt(4x^2+c_{1)},y=-sqrt(4x^2+c_{1)}

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y^'=(4x)/y

y^{^{'}} = \frac{4 x}{y}

y = 4 x^{2} + c_{1}, y = - 4 x^{2} + c_{1}

Solution steps

y^{'} (x) = \frac{4 x}{y}

Solve separable ODE:

y = 4 x^{2} + c_{1}, y = - 4 x^{2} + c_{1}

y = 4 x^{2} + c_{1}, y = - 4 x^{2} + c_{1}

x \to 1 lim ((x) (x - 1)^{- 1})

\int_{0}^{1} x^{3} (1 - x)^{2} d x

\int x e^{- 10 x} d x

\int cos (4 x) d x

y = \frac{2 x}{1 + x ^{2}}, - 4 \leq x \leq 4

What is the solution for y^'=(4x)/y ?
The solution for y^'=(4x)/y is y=sqrt(4x^2+c_{1)},y=-sqrt(4x^2+c_{1)}