What is the solution for yy^'= x/(x^2+1) ?

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

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yy^'= x/(x^2+1)

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} + y = e^{x}

e^{t} t

\int \frac{1 - cos ( x )}{csc ^{2} ( x )} d x

\frac{d}{d x} ((8 x + 3)^{2})

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

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