Question

$F$ is the family of curves in the $XY$ plane., given by $y=x+c{e}^{-x}$, where $c\in R$ is an arbitrary constant. The orthogonal trajectories of the curve of the family $F$ forms the family $F^{\prime }$. Then which of the following statement(s) is\are TRUE for $F^{\prime }$.

• A. $x{e}^y=e^y(y-2)+c$, is general solution for $F^{\prime }$
• B. $x=y-2$ is a particular solution for $F^{\prime }$
• C. If $F^{\prime }$ passes through $(4,0)$, then $F^{\prime }$ is $x=(y-2)+6e^y$
• D. $xe=e(y-2)+c$, is general solution for $F^{\prime }$

Answer 1

Answer: (A), (B)

The correct options are
A $x{e}^y=e^y(y-2)+c$, is general solution for $F^{\prime }$
B $x=y-2$ is a particular solution for $F^{\prime }$

The differential equation of
$F:\frac{dy}{dx}+y=x+1$
The D. E. of orthogonal trajectories of $F$ is
$\frac{-dx}{dy}+y=x+1$
$\frac{dx}{dy}+x=y-1$
So $x.e^{\int 1.dy}=\int e^{\int 1.dy}(y-1)dy+c$
$x{e}^y=e^y(y-2)+c$
Particular solution is $x=y-2$