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

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

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5yy^'=x

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

y = \frac{x ^{2} + c _{1}}{5}, y = - \frac{x ^{2} + c _{1}}{5}

Solution steps

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

Solve separable ODE:

y = \frac{x ^{2} + c _{1}}{5}, y = - \frac{x ^{2} + c _{1}}{5}

y = \frac{x ^{2} + c _{1}}{5}, y = - \frac{x ^{2} + c _{1}}{5}

h \to \infty lim (\frac{5 h ^{4} - 2 h ^{2} + 3}{3 h ^{3} + 2 h ^{2} + h})

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

\frac{d y}{d x} = 2 x (1 + y) e^{x^{2}}

y = 2 x^{2} + 1

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3} + \frac{1}{4 x + 4})

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