What is the solution for 8yy^'=x ?

The solution for 8yy^'=x is y=(sqrt(x^2+c_{1)})/(2sqrt(2)),y=-(sqrt(x^2+c_{1)})/(2sqrt(2))

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8yy^'=x

8 y y^{^{'}} = x

y = \frac{x ^{2} + c _{1}}{2 2}, y = - \frac{x ^{2} + c _{1}}{2 2}

Solution steps

8 y y^{'} (x) = x

Solve separable ODE:

y = \frac{x ^{2} + c _{1}}{2 2}, y = - \frac{x ^{2} + c _{1}}{2 2}

y = \frac{x ^{2} + c _{1}}{2 2}, y = - \frac{x ^{2} + c _{1}}{2 2}

\frac{6}{x ^{5}} - \frac{2}{x}

\frac{\partial}{\partial x} (z^{3} - x^{2} y)

\int (7 - 3 x - 9 x^{2}) d x

\frac{d}{d x} (\frac{2 x ^{2} + 1}{x ^{2} + 7})

What is the solution for 8yy^'=x ?
The solution for 8yy^'=x is y=(sqrt(x^2+c_{1)})/(2sqrt(2)),y=-(sqrt(x^2+c_{1)})/(2sqrt(2))