Question

If the general solution of the differential equation \(y\sec ^{2}\left(\frac{x}{y}\right)dx-\left(x\sec ^{2}\left(\frac{x}{y}\right)-y\right)dy=0\) is \(\tan \left(\frac{x}{y}\right)+g(y)=c\), then \(g(y)=\ldots\)

• A. \(\log |y|\)
• B. \(\tan \sqrt{y}\)
• C. \(\cos 2y\)
• D. \(3y^{2}+1\)

Hint:

Put x equal to v times y and separate the variables, then integrate both sides.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The equation is homogeneous in \(x\) and \(y\), so substitute \(x=vy\).

Step 2: Substitute
Then \(dx=v\,dy+y\,dv\). The equation \(y\sec^{2}\left(\dfrac xy\right)dx=\left(x\sec^{2}\left(\dfrac xy\right)-y\right)dy\) becomes
\(y\sec^{2}v\,(v\,dy+y\,dv)=(vy\sec^{2}v-y)dy\).

Step 3: Separate
\(y^{2}\sec^{2}v\,dv=-y\,dy\), so \(\sec^{2}v\,dv=-\dfrac{dy}{y}\).

Step 4: Integrate
\(\tan v=-\log|y|+c\), so \(\tan\left(\dfrac xy\right)+\log|y|=c\).
Hence \(g(y)=\log|y|\).

Final Answer:
The function g(y) is log of the modulus of y. \[ \boxed{\text{(A) }\log|y|} \]