Question
Solve the differential equation $\frac{d y}{d x}=1-x y+y-x$
✓
Answer
Given, $\frac{d y}{d x}=1-x y+y-x$
= 1 + y - x(1 + y)
$\Rightarrow \quad \frac{d y}{d x}=(1+y)(1-x)$
$\frac{d y}{1+y}=(1-x) d x$
Integrating both sides we get
$\int \frac{d y}{1+y}=\int(1-x) d x$
$\log (1+y)=x-\frac{x^2}{2}+c$
= 1 + y - x(1 + y)
$\Rightarrow \quad \frac{d y}{d x}=(1+y)(1-x)$
$\frac{d y}{1+y}=(1-x) d x$
Integrating both sides we get
$\int \frac{d y}{1+y}=\int(1-x) d x$
$\log (1+y)=x-\frac{x^2}{2}+c$
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