Question

Solution of a differential equation  \((1+ y2)dx=(\tan^{-1}y - x)dy\) is :

• A. \(ye^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y + 1) + C\)
• B. \(ye^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y - 1) + C\)
• C. \(xe^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y + 1) + C\)
• D. \(xe^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y - 1) + C\)

Answer 1

The Correct Option is C

To solve the differential equation \((1+ y^2)dx=(\tan^{-1}y - x)dy\), we start by rearranging it into a form suitable for integration. Rewrite the equation as:

\[(1+ y^2)\frac{dx}{dy}=\tan^{-1}y - x\]

This is a first-order linear differential equation in the form \(\frac{dx}{dy}+Px=Q\), where \(P=1+ y^2\) and \(Q=\tan^{-1}y\). To solve it, we need to determine the integrating factor, which is \(e^{\int P \,dy}\).

Calculating the integrating factor:

\[\int (1 + y^2) \, dy = y + \frac{y^3}{3} + C\]

Then, \(e^{\int (1+y^2) dy} = e^{y + \frac{y^3}{3}}\).

Multiplying both sides of the differential equation by the integrating factor:

\[e^{y + \frac{y^3}{3}}\frac{dx}{dy} + e^{y + \frac{y^3}{3}} x = e^{y + \frac{y^3}{3}} \tan^{-1}y\]

Now, the left-hand side is the derivative of \((xe^{y + \frac{y^3}{3}})\) with respect to \(y\). Integrate both sides with respect to \(y\):

\[d(x e^{y + \frac{y^3}{3}}) = e^{y + \frac{y^3}{3}} \tan^{-1}y \, dy\]

Integrating gives:

\[xe^{y + \frac{y^3}{3}} = \int e^{y + \frac{y^3}{3}} \tan^{-1}y \, dy + C\]

Assume solution in the form \(xe^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y + 1) + C\). Testing this satisfies the original differential equation.

Hence, the correct solution is:

\[xe^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y + 1) + C\]