Question

A differential equation $ \frac{dy}{dx}= \left(y+4x -4\right)^{2} $ is given. At $ x = 0, y = 4 $ the solution of this differential equation is given by

• A. $ y + 4x = 2tan \,(2x) $
• B. $ y + 4x - 4 = 2 \,tan\, (x) $
• C. $ y + 4x - 4 = 2 \,tan\, (2x) $
• D. $ y + 4x - 4 = tan\, (2x) $

Answer 1

The Correct Option is C

We have, $\frac{dy}{dx}=(y+4x-4)^{2}$
put $y+4x=u$ so that $\frac{dt}{dx}+4=\frac{du}{dx}$
So, the given differential equation becomes
$\frac{du}{dx}-4$
$=(u-4)^{2}$
$\Rightarrow \frac{du}{dx}$
$=(u+4)^{2}+4$
$\Rightarrow \frac{du}{\left(u-4\right)^{2}+\left(2\right)^{2}}=dx$
On integrating, we get
$\int \frac{du}{\left(u-4\right)^{2}+\left(2\right)^{2}}=\int dx +c $
$\Rightarrow \frac{1}{2} tan^{-1}\left(\frac{u-4}{2}\right)=x+c$
$\Rightarrow \frac{1}{2}tan^{-1} \left(\frac{y+4x-4}{2}\right)$
$=x+c \left[\because u=y+4x\right]$
When $x = 0$, $y=4$, we have
$\frac{1}{2} tan^{-1}\left(0\right)=0+c$
$\Rightarrow c=0$
$\therefore$ Solution is, $\frac{1}{2} tan^{-1} \left(\frac{y+4x-4}{2}\right)=x$
$\Rightarrow y+4x-4=2$ tan $\left(2x\right)$, which is required solution