What is the solution for 4yy^'=x ?

The solution for 4yy^'=x is y=(sqrt(x^2+c_{1)})/2 ,y=-(sqrt(x^2+c_{1)})/2

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4yy^'=x

4 y y^{^{'}} = x

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

Solution steps

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

Solve separable ODE:

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

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

What is the solution for 4yy^'=x ?
The solution for 4yy^'=x is y=(sqrt(x^2+c_{1)})/2 ,y=-(sqrt(x^2+c_{1)})/2