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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

1 + 4 x - x^{2}, 11 - x - x^{2}, 1 - \frac{2 x}{3}

x + 1, 2 x - 4, 0, 3

\int \frac{3 x ^{5} - 2}{4 x - 4 x ^{5}} d x

\frac{3}{x}, - 1

n = 1 \sum \infty cos (3 n)

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