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}

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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)}