Question

The solution of \(\left(1+y^2\right)+\left(x-\mathrm{e}^{\tan ^{-1} y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0\) is

• A. \(2x\mathrm{e}^{\mathrm{tan}^{-1}y}=\mathrm{e}^{2\mathrm{tan}^{-1}y}+\mathrm{k}\), where k is the constant of integration
• B. \(x\cdot \mathrm{e}^{\mathrm{tan}^{-1}y}=\mathrm{e}^{\mathrm{tan}^{-1}y}+\mathrm{k}\), where k is the constant of integration
• C. \(x\cdot \mathrm{e}^{2\mathrm{tan}^{-1}y}=\mathrm{e}^{\mathrm{tan}^{-1}y}+\mathrm{k}\), where k is the constant of integration
• D. \(x=2+\mathrm{k}\cdot \mathrm{e}^{-\mathrm{tan}^{-1}y}\), where k is the constant of integration

Hint:

Treat x as a function of y, so the equation becomes a linear equation in x.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Rewrite as \(\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{\tan^{-1}y}}{1+y^2}\), a linear equation in \(x\).

Step 2: Integrating factor
\(IF=e^{\int\frac{dy}{1+y^2}}=e^{\tan^{-1}y}\).

Step 3: Solve
\(x\,e^{\tan^{-1}y}=\int\frac{e^{2\tan^{-1}y}}{1+y^2}dy\). Put \(t=\tan^{-1}y\): \(\int e^{2t}dt=\frac{e^{2t}}{2}+C\).

Step 4: Simplify
\(2x\,e^{\tan^{-1}y}=e^{2\tan^{-1}y}+k\).

Final Answer:
The solution is \(2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+k\). \[ \boxed{\text{(A) }2x\,\mathrm{e}^{\tan^{-1}y}=\mathrm{e}^{2\tan^{-1}y}+\mathrm{k}\text{, where k is the constant of integration}} \]