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

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

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y^'=(x+2)(y-1),y(0)=2

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{y} + y e^{c o s (x)} sin (x) d x = 0

cos^{2} (y) sin (x) d x + sin (x) cos (x) d y = 0

\frac{d y}{d x} = 2 x^{4} y, y > 0

\frac{d y}{d x} = 2 x (y^{2} + 1)

\frac{d y}{d x} = 2 x cos^{2} (y), y (0) = \frac{π}{4}

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