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}

\int \frac{x ^{2}}{( 3 + x ^{3} ) ^{2}} d x

\frac{4}{( s ^{2} - 9 )}

\int_{1}^{2} \frac{1}{x} - \frac{1}{x ^{2}} d x

\int \frac{y}{y ^{2} + 2 y + 1} d y

x \to 0 lim (6 x - 1)

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