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} \times3 \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 + 27 y\right) =x+\frac{\pi}{12}$
$\Rightarrow 3x+27y =\tan 3\left(x + \frac{\pi}{12}\right)$