What is the solution for yy^'+25x=0 ?

The solution for yy^'+25x=0 is y=sqrt(-25x^2+c_{1)},y=-sqrt(-25x^2+c_{1)}

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yy^'+25x=0

y y^{^{'}} + 25 x = 0

y = - 25 x^{2} + c_{1}, y = - - 25 x^{2} + c_{1}

Solution steps

y y^{'} (x) + 25 x = 0

Solve separable ODE:

y = - 25 x^{2} + c_{1}, y = - - 25 x^{2} + c_{1}

y = - 25 x^{2} + c_{1}, y = - - 25 x^{2} + c_{1}

\frac{d}{d x} (- 2 x^{- 1})

y^{^{'}} = 2 x - y, y (0) = 1

y^{^{'}} = x^{9} + 1

\frac{\partial}{\partial x} (\frac{( 2 + t ^{2} )}{3 x - x ^{3}})

d er i v a t i v e y = e^{2 - x}

What is the solution for yy^'+25x=0 ?
The solution for yy^'+25x=0 is y=sqrt(-25x^2+c_{1)},y=-sqrt(-25x^2+c_{1)}