Q.Solve the following differential equation: $y \ dx + (x - y^2) \ dy = 0$

Concept: Integrating Factor Method – The equation is not exact, so we find an integrating factor that depends only on $y$.

Rewrite as

This is linear in $x$ with $P(y) = \frac{1}{y}$, $Q(y) = y$.

The integrating factor is

Multiply through: …

The key insight here is to notice which variable is easier to treat as the dependent variable. The equation is given as $y \, dx + (x - y^2) \, dy = 0$. If we try to write it as $\frac{dy}{dx} = \dots$, we get a messy expression that isn't linear. But if we instead treat $x$ as a function of $y$, the structure becomes much cleaner.

Rewrite the equation by dividing through by $dy$ (assuming $dy \neq 0$):

Now rearrange to isolate the derivative term:

Divide through by $y$ (valid for $y \neq 0$):

This is now a first-order linear differential equation in $x$ with respect to $y$. The standard form is $\frac{dx}{dy} + P(y) x = Q(y)$, where $P(y) = \frac{1}{y}$ and $Q(y) = y$.

The integrating factor method works because multiplying the entire equation by $\mu(y)$ turns the left-hand side into the derivative of $\mu(y) x$, which we can then integrate directly.

1. Compute the integrating factor

   $\int P(y) \, dy = \int \frac{1}{y} \, dy = \log|y|$

   So $\mu(y) = e^{\log|y|} = |y|$. For simplicity, we take $\mu(y) = y$ (assuming $y > 0$; the constant $C$ will absorb sign differences later).

2. Multiply the ODE by $\mu(y)$

Notice this is exactly the equation we had before dividing by $y$ — the integrating factor has restored the original left-hand side. The left side is now $\frac{d}{dy}(y x)$.

3. Rewrite and integrate

Integrate both sides with respect to $y$: …

Method: Treat $x$ as a function of $y$ for $M\,dx+N\,dy=0$

When $M\,dx+N\,dy=0$ is not separable or linear in $y$, check whether it is linear in $x$ as a function of $y$.

Steps

Step 1: Divide by $dy$.

Rewrite as $\dfrac{dx}{dy}$ in terms of $x$ and $y$.

Step 2: Look for linear-in-$x$ form.

Rearrange to $\dfrac{dx}{dy}+P(y)\,x=Q(y)$. A term like $y^2$ multiplying $dy$ is a strong hint that $x(y)$ is the clean choice.

Step 3: Integrating factor and integrate. …

Common Mistakes

Mistake 1: Insisting on $\dfrac{dy}{dx}$ form.

Why it's wrong: $y\,dx+(x-y^2)dy=0$ is messy as $y(x)$ but clean as $x(y)$: $\dfrac{dx}{dy}+\tfrac1y x=y$. Correct approach: treat $x$ as a function of $y$.

Mistake 2: Forgetting the $\tfrac{C}{y}$ term. …