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

The solution for y^'=2x(1+x^2-y),y(1)=2 is y=x^2+e^{1-x^2}

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

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

y = x^{2} + e^{1 - x^{2}}

Solution steps

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

Solve linear ODE:

y = x^{2} + e^{1 - x^{2}}

y = x^{2} + e^{1 - x^{2}}

\int cos^{2} (x) ln (x) d x

f (x) = (1 + x + x^{2})^{99}

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

\int \frac{1}{x ^{1.5}} d x

f (x) = 2 cos (sin (x))

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