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

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

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

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

y = e^{x + c_{1}} + e^{x + c_{1}} x + 1

Solution steps

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

Solve separable ODE:

y = e^{x + c_{1}} + e^{x + c_{1}} x + 1

y = e^{x + c_{1}} + e^{x + c_{1}} x + 1

x \to \frac{x}{6} lim (cot (x))

y = x^{2} + 5 x

16 x, \frac{4}{x}, \frac{1}{4} x

\int \frac{sin ( x )}{4 - cos ( x )} d x

\frac{\partial}{\partial x} (\frac{z ^{2} + 2 x + y}{x y})

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