Question:
Which of the following is true for $y(x)$ that satisfies the differential equation $\frac{d y}{d x}=x y-1+x-y ; y(0)=0$
Correct Option: 1
Solution:
$\frac{d y}{d x}=(1+y)(x-1)$
$\frac{d y}{(y+1)}=(x-1) d x$
Integrate $\ln (y+1)=\frac{x^{2}}{2}-x+c$
$(0,0) \Rightarrow \mathrm{c}=0 \Rightarrow \mathrm{y}=\mathrm{e}^{\left(\frac{x^{2}}{2}-\mathrm{x}\right)}-1$
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