What is the solution for y*y^'+36x=0 ?

The solution for y*y^'+36x=0 is y=sqrt(-36x^2+c_{1)},y=-sqrt(-36x^2+c_{1)}

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y*y^'+36x=0

y \cdot y^{^{'}} + 36 x = 0

y = - 36 x^{2} + c_{1}, y = - - 36 x^{2} + c_{1}

Solution steps

y \cdot y^{'} (x) + 36 x = 0

Solve separable ODE:

y = - 36 x^{2} + c_{1}, y = - - 36 x^{2} + c_{1}

y = - 36 x^{2} + c_{1}, y = - - 36 x^{2} + c_{1}

y = x, y = \frac{x}{2}

\frac{d}{d x} (a e^{- 2 x})

\frac{d y}{d x}, 4 x^{2} y^{7} - 2 x = x^{5} + 4 y^{3}

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

f (x) = e^{x} - 1

What is the solution for y*y^'+36x=0 ?
The solution for y*y^'+36x=0 is y=sqrt(-36x^2+c_{1)},y=-sqrt(-36x^2+c_{1)}