Q79 / 120 · VITEEE 2011 · Mathematics
The solution of the differential equation $\frac{d y}{d x}=(4 x+y+1)^{2}$, is
- A$(4 x+y+1)=\tan (2 x+C)$
- B$(4 x+y+1)^{2}=2 \tan (2 x+C)$
- C$(4 x+y+1)^{3}=3 \tan (2 x+$ C) $)$
- Correct answerD$(4 x+y+1)=2 \tan (2 x+\mathrm{C})$
Solution
$$
\begin{array}{l}
\frac{\mathrm{dy}}{\mathrm{dx}}=(4 \mathrm{x}+\mathrm{y}+1)^{2}...(i) \\
\text { Put } 4 \mathrm{x}+\mathrm{y}+1=\mathrm{v} \\
\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}-4 \\
\frac{\mathrm{dv}}{\mathrm{dx}}-4=\mathrm{v}^{2} \\
\Rightarrow \frac{\mathrm{dv}}{\mathrm{v}^{2}+4}=\mathrm{dx} \quad(\because \text { From eq. (i)) } \\
\Rightarrow \frac{1}{2} \tan ^{-1}\left(\frac{\mathrm{v}}{2}\right)=\mathrm{x}+\mathrm{C} \\
\Rightarrow \tan ^{-1}\left(\frac{4 \mathrm{x}+\mathrm{y}+1}{2}\right)=2 \mathrm{x}+\mathrm{C} \\
\Rightarrow 4 \mathrm{x}+\mathrm{y}+1=2 \tan (2 \mathrm{x}+\mathrm{C})
\end{array}
$$
Hence, the correct answer is option: 'D'.