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

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

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(yln(x))y^'=(y^2+1)/x

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{x sin ( x )}{y}, y (0) = - 7

\frac{d y}{d x} = \frac{x sin ( x )}{y}, y (0) = - 9

y^{^{'}} = 2 - 0.1 y

\frac{d x}{d t} = x^{2} + \frac{1}{25}, x (0) = - 3

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

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