The general solution of the differential equation $\frac{d y}{d x}=\cos ^2(3 x+y)$ is $\tan…
The general solution of the differential equation $\frac{d y}{d x}=\cos ^2(3 x+y)$ is $\tan ^{-1}\left(\frac{\sqrt{3}}{2} \tan (3 x+y)\right)=f(x)$. Then, $f(x)=$
$2 \sqrt{3}(x+C)$
$x+C$
$\frac{x+C}{2 \sqrt{3}}$
$\frac{\sqrt{3}}{2}(x+C)$
Solution
Here, $\frac{d y}{d x}=\cos ^2(3 x+y)$
On putting $3 x+y=t$
$3+\frac{d y}{d x}=\frac{d t}{d x}$
$\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=\frac{d t}{d x}-3 \Rightarrow \frac{d t}{d x}-3=\cos ^2 t \\ \Rightarrow & \frac{d t}{d x}=\cos ^2 t+3 \Rightarrow \frac{d t}{\cos ^2 t+3}=d x\end{array}$
Integrate both side,
$\int \frac{d t}{\cos ^2 t+3}=\int d x$
$\begin{aligned} & \Rightarrow \int \frac{\sec ^2 t d t}{1+3 \sec ^2 t}=\int d x \Rightarrow \int \frac{\sec ^2 t}{1+3+3 \tan ^2 t}=\int d x \\ & \Rightarrow \int \frac{\sec ^2 t d t}{4+3 \tan ^2 t}=\int d x\end{aligned}$
On putting $\tan t=m$,
$\sec ^2 t d x=d m=\frac{1}{4} \int \frac{d m}{1+\left(\frac{\sqrt{3}}{2} m\right)^2} x+C$
$=\frac{1}{4} \times \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{2} m\right)=x+C$
$=\frac{1}{2 \sqrt{3}} \tan ^{-1}\left[\frac{\sqrt{3}}{2} \tan t\right]=x+C$ $[\because m=\tan t]$
$=\tan ^{-1}\left[\frac{\sqrt{3}}{2} \tan (3 x+y)=2 \sqrt{3}(x+C)\right]$ $[\therefore t=3 x+y]$
So, $f(x)=2 \sqrt{3}(x+C)$