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: Formula First
For $\tan\left(\dfrac xy\right)$: $d\left[\tan\dfrac xy\right]=\sec^{2}\left(\dfrac xy\right)\dfrac{y\,dx-x\,dy}{y^{2}}$.

Step 2: Rearrange the Equation
The given equation is $\sec^{2}\left(\dfrac xy\right)(y\,dx-x\,dy)+y\,dy=0$.

Step 3: Divide by y Squared
$\sec^{2}\left(\dfrac xy\right)\dfrac{y\,dx-x\,dy}{y^{2}}+\dfrac{dy}{y}=0$, which is $d\left[\tan\dfrac xy\right]+d\left[\log|y|\right]=0$.

Step 4: Conclude
Integrating gives $\tan\left(\dfrac xy\right)+\log|y|=c$, so $g(y)=\log|y|$.

Final Answer:
Matching the differential gives g of y equal to log of the modulus of y. \[ \boxed{\text{(A) }\log|y|} \]