What is the solution for (1+y^2)dx-(x+x^2)dy=0 ?

The solution for (1+y^2)dx-(x+x^2)dy=0 is y=tan(ln(x)-ln(x+1)-c_{1})

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(1+y^2)dx-(x+x^2)dy=0

(1 + y^{2}) d x - (x + x^{2}) d y = 0

y = tan (ln (x) - ln (x + 1) - c_{1})

Solution steps

(1 + y^{2}) d x - (x + x^{2}) d y = 0

Solve separable ODE:

y = tan (ln (x) - ln (x + 1) - c_{1})

y = tan (ln (x) - ln (x + 1) - c_{1})

\frac{x ^{2}}{1 + x ^{3}} d x - \frac{1}{y} d y = 0

\frac{d y}{d x} = - \frac{x}{y}, y (1) = - 2

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

t y^{^{'}} + 2 y = \frac{1}{y ^{2}}

\frac{d v}{d t} = 9 v

What is the solution for (1+y^2)dx-(x+x^2)dy=0 ?
The solution for (1+y^2)dx-(x+x^2)dy=0 is y=tan(ln(x)-ln(x+1)-c_{1})