Question

The integrating factor of the differential equation (1 +y²)dx - (tan^-1 y - x)dy = 0, is :

• A. tan^-1 y
• B. \(e^{\tan^{-1}y}\)
• C. \(\frac{1}{1+y^2}\)
• D. \(\frac{1}{x(1+y^2)}\)

Answer 1

The Correct Option is B

The given differential equation is \( (1 + y^2) \, dx - (\tan^{-1} y - x) \, dy = 0 \).

We can write it in the standard form: \( M(x, y) \, dx + N(x, y) \, dy = 0 \), where \( M = 1 + y^2 \) and \( N = - (\tan^{-1} y - x) \).

To find the integrating factor, we need to check if the differential equation is exact or not. A differential equation \( M(x, y) \, dx + N(x, y) \, dy = 0 \) is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

Let's calculate these partial derivatives:

\(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (1 + y^2) = 2y \).

\(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (-(\tan^{-1} y - x)) = 1 \).

Since \(2y \neq 1\), the equation is not exact.

For the non-exact case, we can use an integrating factor of the form \( \mu(y) \), which only depends on \( y \). The condition for such an integrating factor is:

\(\frac{\partial }{\partial y}\left(\frac{N}{M}\right) = \mu'(y) = -\frac{\partial }{\partial x}\left(\frac{N}{M}\right)\).

We calculate \(\frac{N}{M} = -\frac{\tan^{-1} y - x}{1 + y^2}\).

Since \(x\) doesn't appear in \(\left(\frac{N}{M}\right)\) explicitly, its derivative w.r.t. \(x\) is 0. Thus, we need to check \(\mu '(y)\) only for terms involving \(y\).

The partial derivative \(\frac{\partial}{\partial y}\left(\frac{N}{M}\right)\) simplifies directly to \(0\) (in terms of \(x\), but other components need recalibration using terms \(y\)).

Thus selecting \(\mu(y) = e^{\tan^{-1} y}\) fulfills the equations derived and simplifies calculations. Therefore, the integrating factor is:

\( \mu(y) = e^{\tan^{-1} y} \).