Question

The general solution of the differential equation $(x + y + 3) \,\frac{dy}{dx}\, =\,1$ is

• A. $x + y + 3 = Ce^y$
• B. $x + y + 4 = Ce^y$
• C. $x + y + 3 = Ce^{-y}$
• D. $x + y + 4 = Ce^{-y}$
• E. $x + y + 4e^y = C$

Answer 1

Answer: (B)

We have, $(x+y+3) \frac{d y}{d x}=1$
$\Rightarrow (x+y+3)=\frac{d x}{d y} $
Let $x+y+3=t$
On differentiating w.r.t.y, we get
$\frac{d x}{d y}+1=\frac{d t}{d y}$
$\Rightarrow \frac{d t}{d y}=t+1\left[\because \text { from Eq. (i), } t=\frac{d x}{d y}\right]$
On integrating both sides,
$\int \frac{d t}{(t+1)}=\int d y$
$\Rightarrow \log (t+1)=y+C_{1}$
$\Rightarrow \log (x+y+3+1)=y+C_{1}$
$\therefore x+y+4=C e^{y}$ [where ,$c=e^{c_{1}}$]