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

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

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x+yy^'=0

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

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

Solution steps

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

Solve separable ODE:

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

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

y = x^{3} + 7 x^{2} - 4

\frac{d y}{d t} = 4 t - 0.1 y t

x^{2} - 5 x + 1

- \frac{2 x ^{2}}{5 y ^{2}}

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

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