What is a solution to the differential equation $\frac{d y}{d x} = (1 + x) (1 + y)$ $dy/dx=(1+x)(1+y)$ ?

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Answer 1 · 2016-07-18T14:47:32

$y = C e^{x + \frac{x^{2}}{2}} - 1$ $y =C e^(x + x^2 / 2 ) - 1$

$\frac{d y}{d x} = (1 + x) (1 + y)$ $dy/dx=(1+x)(1+y)$

this is separable!

$\frac{1}{1 + y} \frac{d y}{d x} = 1 + x$ $1/(1+y) dy/dx=1+x$

so we integrate both sides

$int \ 1/(1+y) dy/dx \ dx=int \ 1+x \ dx$

or

$int \ 1/(1+y) \ dy=int \ 1+x \ dx$

$ln (1+y) = x + x^2 / 2 + C$

$1+y = e^(x + x^2 / 2 + C)$

$y =C e^(x + x^2 / 2 ) - 1$

What is a solution to the differential equation dy/dx=(1+x)(1+y)dydx=(1+x)(1+y)dy/dx=(1+x)(1+y)?