Question

The solution of the equation $\log \frac{d y}{d x}=9 x-6 y+6$, given that $y=1$ when $x=0$, is

• A. $3 e^{6 y}=2 e^{9 x-6}+6 e^{6}$
• B. $3 e^{6 y}=2 e^{9 x+6}-6 e^{6}$
• C. $3 e^{6 y}=2 e^{9 x+6}+e^{6}$
• D. None of these

Answer 1

Answer: (C)

We have,
$\log \frac{d y}{d x} =9 x-6 y+6$
$\Rightarrow \frac{d y}{d x}=e^{9 x-6 y+6}=e^{9 x+6} \cdot e^{-6 y} $
$\Rightarrow e^{6 y} d y=e^{9 x+6} d x$
Integrating, we get $\frac{e^{6 y}}{6}=\frac{e^{9 x+6}}{9}+c$
Putting $x=0, y=1$, we get
$\frac{e^{6}}{6}=\frac{e^{6}}{9}+c$ i.e. $c=\frac{e^{6}}{18}$
$\therefore $ The solution is $\frac{e^{6 y}}{6}=\frac{e^{9 x+6}}{9}+\frac{e^{6}}{18}$
$\Rightarrow 3 e^{6 y}=2 e^{9 x+6}+e^{6}$