Question

A particular solution of $\frac{dy}{dx}=(x+9y{)}^2$ when $x=0,y=\frac{1}{27}$ is

• A. $3x+27y=\tan 3\left(x+\frac{\pi }{12}\right)$
• B. $3x+27y={\tan }^{-1}3\left(x+\frac{\pi }{12}\right)$
• C. $3x+27y=\tan 9\left(x+\frac{\pi }{12}\right)$
• D. $3x+27y=\tan \left(x+\frac{\pi }{12}\right)$

Answer 1

Answer: (A)

We have , $\frac{dy}{dx}={\left(x+9y\right)}^2$ ....(i)
Put $x+9y=t\Rightarrow 1+9\frac{dy}{dx}=\frac{dt}{dx}$
Now (i) becomes
$\Rightarrow \frac{dt}{dx}-1=9t^2\Rightarrow \frac{dt}{9t^2+1}=dx$
$\Rightarrow \frac{1}{9}\int \frac{dt}{t^2+\frac{1}{9}}=\int dx+C$
$\Rightarrow \frac{1}{9}\times 3{\tan }^{-1}\left(3t\right)=x+C$
$\Rightarrow \frac{1}{3}{\tan }^{-1}\left(3x+27y\right)=x+C$ .....(ii)
Now, $x=0,y=\frac{1}{27}$
$\Rightarrow \frac{1}{3}{\tan }^{-1}\left(1\right)=0+C\Rightarrow C=\frac{\pi }{12}$
So (ii) becomes
$\therefore \frac{1}{3}{\tan }^{-1}\left(3x+27y\right)=x+\frac{\pi }{12}$
$\Rightarrow 3x+27y=\tan 3\left(x+\frac{\pi }{12}\right)$