Q.The Integrating Factor of the differential equation $(1 - y^2)\dfrac{dx}{dy} + yx = ay \ (-1 < y < 1)$ is (A) $\dfrac{1}{y^2 - 1}$ (B) $\dfrac{1}{\sqrt{y^2 - 1}}$ (C) $\dfrac{1}{1 - y^2}$ (D) $\dfrac{1}{\sqrt{1 - y^2}}$

The key idea is to rewrite the equation in the standard linear form $\frac{dx}{dy} + P(y)x = Q(y)$ and then compute the integrating factor $\mu(y) = e^{\int P(y)\,dy}$.

Step 1: Divide the given equation by $(1 - y^2)$:

Here $P(y) = \frac{y}{1 - y^2}$.

Step 2: Compute $\int P(y)\,dy$:

$\int \frac{y}{1 - y^2}\,dy = -\frac12 \log|1 - y^2| = \log\left(\frac{1}{\sqrt{1 - y^2}}\right),$ …

The Integrating Factor (IF) method is the standard tool for solving first-order linear differential equations. The idea is simple: if you have an equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, you can multiply both sides by a cleverly chosen function $\mu(y)$ — the integrating factor — so that the left-hand side becomes the exact derivative of $\mu(y) \cdot x$. That turns the problem into a straightforward integration.

Here, the equation is given as $(1 - y^2)\frac{dx}{dy} + yx = ay$. Notice the independent variable is $y$, not $x$ — that’s fine. We just need to get it into the standard linear form with $\frac{dx}{dy}$ alone.

1. Divide through by the coefficient of $\frac{dx}{dy}$. The coefficient is $(1 - y^2)$. Since $-1 < y < 1$, $1 - y^2 > 0$, so division is safe.

Now it matches $\frac{dx}{dy} + P(y)x = Q(y)$ with $P(y) = \frac{y}{1 - y^2}$.

2. Find the integrating factor $\mu(y)$. The formula is $\mu(y) = e^{\int P(y)\,dy}$. So we need:

This is a standard substitution: let $u = 1 - y^2$, then $du = -2y\,dy$, so $y\,dy = -\frac{1}{2}du$.

Since $1 - y^2 > 0$ in the given domain, we can drop the absolute value.

3. Exponentiate to get $\mu(y)$.

The constant of integration is irrelevant here — we only need one integrating factor, so we take the simplest form. …

Method: Integrating factor when the equation is linear in $x(y)$

When the derivative is $\dfrac{dx}{dy}$, the integrating factor uses an integral in $y$; compute it carefully.

Steps

Step 1: Divide by the coefficient of $\dfrac{dx}{dy}$.

Bring it to $\dfrac{dx}{dy}+P(y)\,x=Q(y)$; here $P(y)=\dfrac{y}{1-y^2}$.

Step 2: Integrate $P(y)$ by substitution.

With $u=1-y^2$, $\displaystyle\int\frac{y}{1-y^2}\,dy=-\tfrac12\log|1-y^2|$. …

Common Mistakes

Mistake 1: Not dividing by $(1-y^2)$ before reading $P(y)$.

Why it's wrong: standard form needs coefficient $1$ on $\dfrac{dx}{dy}$, giving $P(y)=\dfrac{y}{1-y^2}$. Correct approach: divide the whole equation by $1-y^2$ first.

Mistake 2: Dropping the minus sign in $\int\dfrac{y}{1-y^2}\,dy$. …