What is the solution for xyy^'=x^2+1 ?

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

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xyy^'=x^2+1

x y y^{^{'}} = x^{2} + 1

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

Solution steps

x y y^{'} (x) = x^{2} + 1

Solve separable ODE:

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

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

\int 4 x ln (x) d x

x^{2} - x y + y^{2} = 3, (1, 2)

\frac{4 x ^{5} + 5}{x ^{3}}

y^{^{'}} = x y^{2}

\frac{d}{d x} (e^{x} sin (x) cos (x))

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