Question:
If the general solution of the differential equation \(\frac{dy}{dx}=\cos (2x-3y)-\cos (2x+3y)\) is \(\cos 2x=g(y)+c\), then \(g(y)=\ldots\)
If the general solution of the differential equation \(\frac{dy}{dx}=\cos (2x-3y)-\cos (2x+3y)\) is \(\cos 2x=g(y)+c\), then \(g(y)=\ldots\)
Show Hint
Use cos A - cos B = 2 sin((A+B)/2) sin((B-A)/2) to separate the variables, then integrate.
Updated On: Nov 12, 2026
- \(\cos 2x+\cos 2y\)
- \(\sec y+\tan y\)
- \(-\frac{1}{3}\log |\tan \left(\frac{3y}{2}\right)|\)
- \(2\,\log |\text{cosec}3y-\cot 3y|\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Convert the difference of cosines into a product to separate x and y.
Step 2: Simplify
\(\cos(2x-3y)-\cos(2x+3y)=2\sin 2x\sin 3y\).
So \(\frac{dy}{dx}=2\sin 2x\sin 3y\).
Step 3: Separate the variables
\(\frac{dy}{\sin 3y}=2\sin 2x\,dx\Rightarrow\text{cosec}\,3y\,dy=2\sin 2x\,dx\).
Step 4: Integrate
\(\frac{1}{3}\log\left|\tan\frac{3y}{2}\right|=-\cos 2x+c\).
So \(\cos 2x=-\frac{1}{3}\log\left|\tan\frac{3y}{2}\right|+c\).
Final Answer:
g(y) is minus one third of log of the absolute value of tan(3y/2). \[ \boxed{\text{(C) }-\dfrac{1}{3}\log \left|\tan \left(\dfrac{3y}{2}\right)\right|} \]
Convert the difference of cosines into a product to separate x and y.
Step 2: Simplify
\(\cos(2x-3y)-\cos(2x+3y)=2\sin 2x\sin 3y\).
So \(\frac{dy}{dx}=2\sin 2x\sin 3y\).
Step 3: Separate the variables
\(\frac{dy}{\sin 3y}=2\sin 2x\,dx\Rightarrow\text{cosec}\,3y\,dy=2\sin 2x\,dx\).
Step 4: Integrate
\(\frac{1}{3}\log\left|\tan\frac{3y}{2}\right|=-\cos 2x+c\).
So \(\cos 2x=-\frac{1}{3}\log\left|\tan\frac{3y}{2}\right|+c\).
Final Answer:
g(y) is minus one third of log of the absolute value of tan(3y/2). \[ \boxed{\text{(C) }-\dfrac{1}{3}\log \left|\tan \left(\dfrac{3y}{2}\right)\right|} \]
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