Question

The general solution of the differential equation \[ \frac{dy}{dx}=\cos^2(3x+y) \] is \[ \tan^{-1}\left(\frac{\sqrt{3}}{2}\tan(3x+y)\right)=f(x). \] Then \(f(x)=\)

• A. \(2\sqrt{3}(x+C)\)
• B. \(x+C\)
• C. \(\frac{x+C}{2\sqrt{3}}\)
• D. \(\frac{\sqrt{3}}{2}(x+C)\)

Hint:

For differential equations involving expressions like \(ax+y\), use the substitution \[ u=ax+y \] to reduce the equation into separable form.

Answer 1

The Correct Option is A

Step 1: Introduce a substitution.
Given, \[ \frac{dy}{dx}=\cos^2(3x+y) \] Let \[ u=3x+y \] Then, \[ \frac{du}{dx}=3+\frac{dy}{dx} \] Using the differential equation, \[ \frac{du}{dx}=3+\cos^2 u \] Hence, \[ \frac{du}{3+\cos^2 u}=dx \]

Step 2: Simplify the denominator.
Using \[ \cos^2 u=\frac{1}{1+\tan^2 u}, \] let \[ t=\tan u \] Then, \[ du=\frac{dt}{1+t^2} \] Now, \[ 3+\cos^2 u = 3+\frac{1}{1+t^2} \] \[ = \frac{3(1+t^2)+1}{1+t^2} \] \[ = \frac{4+3t^2}{1+t^2} \] Therefore, \[ \frac{du}{3+\cos^2 u} = \frac{\frac{dt}{1+t^2}}{\frac{4+3t^2}{1+t^2}} \] \[ = \frac{dt}{4+3t^2} \] Hence, \[ \int \frac{dt}{4+3t^2}=\int dx \]

Step 3: Integrate both sides.
We know that \[ \int \frac{dt}{a^2+b^2t^2} = \frac{1}{ab}\tan^{-1}\left(\frac{bt}{a}\right)+C \] Here, \[ a=2, \qquad b=\sqrt{3} \] Thus, \[ \int \frac{dt}{4+3t^2} = \frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{2}t\right) \] Therefore, \[ \frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{2}\tan u\right) = x+C \] Multiplying by \(2\sqrt{3}\), \[ \tan^{-1}\left(\frac{\sqrt{3}}{2}\tan u\right) = 2\sqrt{3}(x+C) \] Substituting \[ u=3x+y, \] we get \[ \tan^{-1}\left(\frac{\sqrt{3}}{2}\tan(3x+y)\right) = 2\sqrt{3}(x+C) \] Thus, \[ f(x)=2\sqrt{3}(x+C) \]

Step 4: Final conclusion.
Therefore, \[ \boxed{2\sqrt{3}(x+C)} \]