Question

The solution of $ \frac{dy}{dx}=\left(y+4x+1\right)^{2} $ when $ y(0) = 1 $ , is given by

• A. $ y=2tan\left(2x+\pi/4\right)-4x-1 $
• B. $ y=2cot\left(2x+\pi/8\right)-4x-1 $
• C. $ y=2tan\left(x+\pi/4\right)-4x-1 $
• D. $ y=2cot\left(2x+\pi/4\right)-4x-1 $

Answer 1

Answer: (A)

The given differential equation is,
$\frac{dy}{dx} = (y +4x +1)^2\,\,\,...(i)$
Put $(y + 4x +1 ) = t$
$\Rightarrow \frac{dy}{dx} + 4 = \frac{dt}{dx}$
$\therefore (i)$ becomes,
$\frac{dt}{dx} - 4 = t^2$
$\Rightarrow \frac{dt}{dx} = t^2 + 2^2$
$\Rightarrow \int \frac{dt}{t^2 +2^2} = \int dx$
$\Rightarrow \frac{1}{2} tan^{-1} (\frac{t}{2}) = x + C$
$\Rightarrow \frac{1}{2} tan^{-1} (\frac{y+4x +1}{2}) = x + C$
Now, at $ x= 0, y = 1$
$\therefore \frac{1}{2} tan^{-1} (1) = C $
$\Rightarrow C = \frac{\pi}{8}$
So, the required solution is ,
$\frac{1}{2} tan^{-1} (\frac{y+4x +1}{2}) = x + \frac{\pi}{8}$
$\Rightarrow y + 4x +1 = 2 \,tan(2x + \frac{\pi}{4})$
$\Rightarrow y = 2\,tan (2x + \frac{\pi}{4}) - 4x -1$