Question

If $\frac{dy}{dx}=\frac{(x-1{)}^2+(y+3{)}^2\times e^{\left\{\right(y+3\left)/\right(x-1\left)\right\}}}{(xy+3x-y-3)\times e^{(y+3)/(x-1)}}$ then

• A. $(y-x+4)e^{(y+3)/(x-1)}=(x-1)log|x-1|+C(x-1)$
• B. $(y-x+4)e^{(y+3)/(x-1)}=(x-1)log|x-1|+C$
• C. $(y-x)e^{(y+3)/(x-1)}=(x-1)log|x-1|+(x-1)$
• D. $(y-x-4)e^{(y+3)/(x-1)}=(x-1)log|x-2|+C(x-2)$

Answer 1

The correct option is A $(y-x+4)e^{(y+3)/(x-1)}=(x-1)log|x-1|+C(x-1)$

Given $\frac{dy}{dx}=\frac{(x-1{)}^2+(y+3{)}^2{e}^{\left\{\right(y+3\left)/\right(x-1\left)\right\}}}{(xy+3x-y-3)e^{(y+3)/(x-1)}}$

$\frac{dy}{dx}=\frac{(x-1{)}^2+(y+3{)}^2{e}^{\left\{\right(y+3\left)/\right(x-1\left)\right\}}}{(x-1)(y+3)e^{(y+3)/(x-1)}}$ ..........(1)

Let $y+3=t(x-1)$

Differentiating w.r.t. x, we get,

$\frac{dy}{dx}=(x-1)\frac{dt}{dx}+t$

Thus (1) becomes

$(x-1)\frac{dt}{dx}+t=\left\{\right(1+t^2{e}^t\left)/t{e}^t\right\}$

$\Rightarrow (x-1)\frac{dt}{dx}=\frac{1}{t{e}^t}$

$\Rightarrow t{e}^{t}dt=\frac{dx}{x-1}$

$\Rightarrow t{e}^t-e^t=\log |x-1|+C$

$\Rightarrow (y-x+4)e^{\left\{\right(y+3\left)/\right(x-1\left)\right\}}=(x-1)\log \left|\right(x-1\left)\right|+C(x-1)$