Question

The general solution of the differential equation $\frac{dy}{dx}=(4x+y+1{)}^2$ is (where $c$ is a constant of integration)

• A. $\frac{1}{2}{\tan }^{-1}\left(\frac{4x+y+1}{4}\right)=x+c$
• B. $\frac{1}{2}{\tan }^{-1}\left(\frac{4x+y+1}{2}\right)=x+c$
• C. $\frac{1}{3}{\tan }^{-1}\left(\frac{4x+y+1}{4}\right)=2x+c$
• D. $\frac{1}{4}{\tan }^{-1}\left(\frac{4x+y+1}{2}\right)=|x|+c$

Answer 1

The correct option is B $\frac{1}{2}{\tan }^{-1}\left(\frac{4x+y+1}{2}\right)=x+c$

Putting $4x+y+1=t\Rightarrow 4+\frac{dy}{dx}=\frac{dt}{dx}\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-4$

Given equation becomes

$\frac{dt}{dx}-4=t^2\Rightarrow \frac{dt}{t^2+4}=dx$(Variables are separated)

Integrating both sides,we get

$\underset{}{\overset{}{\int }}\frac{dt}{4+t^2}=\underset{}{\overset{}{\int }}dx\Rightarrow \frac{1}{2}{\tan }^{-1}\frac{t}{2}=x+c$

$\Rightarrow \frac{1}{2}{\tan }^{-1}\left(\frac{4x+y+1}{2}\right)=x+c$