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

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

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yy^'+36x=0

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

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

Solution steps

y 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}

\int \frac{20 x ^{3} - 40 x ^{2} + 9}{x ^{2} - 2 x} d x

\frac{\partial}{\partial z} (e^{x})

y^{3} - (6 x + 2) + 3 x y^{2} y^{^{'}} = 0

t an g e n t (x^{3} - 4 x)^{8}

\frac{\partial}{\partial x} (x^{2} + y^{2} - 1)

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