What is the solution for yy^'=-5x ?

The solution for yy^'=-5x is y=sqrt(-5x^2+c_{1)},y=-sqrt(-5x^2+c_{1)}

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yy^'=-5x

y y^{^{'}} = - 5 x

y = - 5 x^{2} + c_{1}, y = - - 5 x^{2} + c_{1}

Solution steps

y y^{'} (x) = - 5 x

Solve separable ODE:

y = - 5 x^{2} + c_{1}, y = - - 5 x^{2} + c_{1}

y = - 5 x^{2} + c_{1}, y = - - 5 x^{2} + c_{1}

\frac{d y}{d x} - \frac{2 x}{x ^{2} + 1} y = - \frac{2 x}{x ^{2} + 1}

(- s^{2} - s) y^{^{'}} (s) + (- 3 s - 1) y (s) = 0

x d y + y d x = 2 x^{2} 2 y d x

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

(y^{2} + 3) d x = y sec^{2} (x) d y

What is the solution for yy^'=-5x ?
The solution for yy^'=-5x is y=sqrt(-5x^2+c_{1)},y=-sqrt(-5x^2+c_{1)}