Question

The orthogonal trajectory for the family of curves $x+1=y+k{e}^{-y}$ is

• A. $1+y=x+C{e}^{-x}$
• B. $1-y=x+C{e}^{-x}$
• C. $1-y=x-C{e}^{-x}$
• D. None of the above

Answer 1

The correct option is A $1+y=x+C{e}^{-x}$

We have, $x+1=y+k{e}^{-y}$ ....(1)

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

$1=(1-k{e}^{-y})\frac{dy}{dx}$ ....(2)

Elimination $k$ from (2) using (1), we get,

$1=(y-x)\frac{dy}{dx}$ ....(3)

Replacing $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ in (3), we get

$1=-(y-x)\frac{dx}{dy}$

$\Rightarrow \frac{dy}{dx}=x-y$ ....(4)

Substituting $x-y=t$, we get

$1-\frac{dt}{dx}=t⟹\frac{1}{1-t}dt=dx$

Integrating, we get

$-\log (1-t)=x+c⟹1-t=C{e}^{-x}$

$\Rightarrow 1-x+y=C{e}^{-x}⟹1+y=x+C{e}^{-x}$