What is the solution for y^'=1+x-y-xy,y(0)=-1 ?

The solution for y^'=1+x-y-xy,y(0)=-1 is y=-2e^{-(x^2)/2-x}+1

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y^'=1+x-y-xy,y(0)=-1

y^{^{'}} = 1 + x - y - x y, y (0) = - 1

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

Solution steps

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

Solve separable ODE:

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

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

x \to π lim (- 4 sin (\frac{x}{2}))

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

\frac{\partial}{\partial x} (25 x^{8} y^{4} + 12 x^{4} y^{5})

\int 5 e^{x} + cos (\frac{x}{2}) d x

2^{a r c t a n (x)}

What is the solution for y^'=1+x-y-xy,y(0)=-1 ?
The solution for y^'=1+x-y-xy,y(0)=-1 is y=-2e^{-(x^2)/2-x}+1