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 curves of family F form the family $F^1$. Then which of the following statements are true for $F^1$.

• A. $x{e}^y=e^y(y-2)+c$, is general solution for $F^1$.
• B. $x=y-2$ is a particular solution for $F^1$.
• C. If $F^1$ passess through (4, 0), then $F^1$ is $x=(y-2)+6e^y$.
• D. $xe=e(y-2)+c$, is general solution for $F^1$.

Answer 1

Answer: (A), (B)

The correct options are
A $x{e}^y=e^y(y-2)+c$, is general solution for $F^1$.
B $x=y-2$ is a particular solution for $F^1$.

Differential equation of
$F:\frac{dy}{dx}+y=x+1$
now replace $\frac{dy}{dx}\to -\frac{dx}{dy}$
$-\frac{dx}{dy}+y=x+1$
i.e. $\frac{dx}{dy}+x=y-1$, which is linear in x.
So its solution is
$x.e^{\int 1.dy}=\int e^{\int 1.dy}.(y-1)dy+c$
i.e. $x{e}^y=\int e^y(y-1)dy+C$
$x{e}^y=e^y(y-2)+C$
The particular solution is $x=y-2$.