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

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

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(1+x^2)y^'=xy+x^2y^2

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

y = \frac{2 x ^{2} + 1}{- x x ^{2} + 1 + ln ( x ^{2} + 1 + x ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 x ^{2} + 1}{- x x ^{2} + 1 + ln ( x ^{2} + 1 + x ) + c _{1}}

y = \frac{2 x ^{2} + 1}{- x x ^{2} + 1 + ln ( x ^{2} + 1 + x ) + c _{1}}

f (x) = \frac{1}{7 x}, at x = 9

y^{^{''}} - 2 x y^{^{'}} = 0

\int_{- 2}^{62} 2 π (y) (2) d y

\int \frac{2}{x ^{2} - 4 x} d x

sin (2 t) \cdot e^{3 t}

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