Q.Find the general solution of the differential equation $y\,dx - (x + 2y^2)\,dy = 0$.

The key idea is to treat $y$ as the independent variable — the equation is linear in $x$ as a function of $y$.

Rewrite the given equation as:

This is a first-order linear ODE in $x(y)$. The integrating factor is:

Multiply through: …

Why this approach works

When you see a differential equation like $y\,dx - (x + 2y^2)\,dy = 0$, your first instinct might be to rearrange it into the standard form $\frac{dy}{dx} = \dots$ But look carefully: the term $2y^2$ depends only on $y$, and the $x$ appears linearly. This is a strong hint that treating $x$ as the dependent variable (a function of $y$) will be much cleaner.

The equation is linear in $x$ but not in $y$. By writing it as $\frac{dx}{dy} = \dots$, we get a first-order linear ODE — and those have a standard, reliable solution method using an integrating factor.

Step-by-step solution

1. Rewrite the equation in standard linear form

Start with:

Bring the $dy$ term to the other side:

Divide through by $y\,dy$ (assuming $y \neq 0$):

Separate the fraction:

Now bring the $x/y$ term to the left:

This is now in the standard form for a linear first-order ODE in $x(y)$:

where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y$.

2. Find the integrating factor

For a linear ODE $\frac{dx}{dy} + P(y)\,x = Q(y)$, the integrating factor is:

Here $P(y) = -\frac{1}{y}$, so:

Therefore:

Since we typically work with a particular solution, we can take the integrating factor as $\mu(y) = \frac{1}{y}$ (the sign will be absorbed by the constant later).

Method: Linear in $x$ — swap the roles of the variables

Use this when writing $\dfrac{dy}{dx}$ gives a non-linear mess, but treating $x$ as a function of $y$ makes the equation linear. A tell is that $x$ appears only to the first power while $y$ appears in awkward ways.

Steps

Step 1: Rearrange to $\dfrac{dx}{dy} + P(y)\,x = Q(y)$.

Divide by whatever multiplies $dx$ and by $dy$, and collect the $x$-term on the left. Now $x$ is the dependent variable and $y$ the independent one.

Step 2: Integrating factor in $y$.

$\mu(y) = e^{\int P(y)\,dy}.$ …

Common Mistakes

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

Why it's wrong: written as $\dfrac{dy}{dx}$ the equation is not linear in $y$, so the standard method stalls. Correct approach: notice $x$ appears linearly and switch to $\dfrac{dx}{dy}$, treating $x$ as the dependent variable.

Mistake 2: Sign error forming the standard form. …