Question

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

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

Answer 1

Answer: (C)

We have,
$\log \frac{dy}{dx}=9x-6y+6$
$\Rightarrow \frac{dy}{dx}=e^{9x-6y+6}=e^{9x+6}\cdot e^{-6y}$
$\Rightarrow e^{6y}dy=e^{9x+6}dx$
Integrating, we get $\frac{e^{6y}}{6}=\frac{e^{9x+6}}{9}+c$
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 $\frac{e^{6y}}{6}=\frac{e^{9x+6}}{9}+\frac{e^6}{18}$
$\Rightarrow 3e^{6y}=2e^{9x+6}+e^6$