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,
$\frac{dt}{dx}-4=t^2$
$\Rightarrow \frac{dt}{dx}=t^2+2^2$
$\Rightarrow \int \frac{dt}{t^2+2^2}=\int dx$
$\Rightarrow \frac{1}{2}ta{n}^{-1}(\frac{t}{2})=x+C$
$\Rightarrow \frac{1}{2}ta{n}^{-1}(\frac{y+4x+1}{2})=x+C$
Now, at $x=0,y=1$
$\therefore \frac{1}{2}ta{n}^{-1}(1)=C$
$\Rightarrow C=\frac{\pi }{8}$
So, the required solution is ,
$\frac{1}{2}ta{n}^{-1}(\frac{y+4x+1}{2})=x+\frac{\pi }{8}$
$\Rightarrow y+4x+1=2tan(2x+\frac{\pi }{4})$
$\Rightarrow y=2tan(2x+\frac{\pi }{4})-4x-1$