Q.Find the general solution of the differential equation $x\frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$.

The key idea is that this is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Step 1: Rewrite in standard form

Divide through by $x$:

Here $P(x) = \frac{2}{x}$ and $Q(x) = x$.

Step 2: Find the integrating factor

Step 3: Multiply through and integrate

Multiplying the standard form by $x^2$:

Integrate both sides:

Step 4: Solve for $y$

The equation $x\frac{dy}{dx} + 2y = x^2$ is a first-order linear differential equation. The key idea: we can rewrite it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ and then multiply both sides by an integrating factor — a function that turns the left-hand side into a perfect derivative of a product. This trick works because the derivative of a product always gives two terms, just like the left side of our equation.

Let’s walk through it.

1. Rewrite in standard form. Divide through by $x$ (allowed since $x \neq 0$):

Here $P(x) = \frac{2}{x}$ and $Q(x) = x$.

2. Find the integrating factor. The integrating factor $\mu(x)$ is given by $e^{\int P(x)\,dx}$.

So $\mu(x) = e^{\log(x^2)} = x^2$.

3. Multiply the ODE by $\mu(x)$.

Simplify:

4. Recognise the left side as a derivative. Notice that $\frac{d}{dx}(x^2 y) = x^2\frac{dy}{dx} + 2xy$. Exactly our left side! So the equation becomes:

5. Integrate both sides.

where $C$ is the constant of integration.

6. Solve for $y$. Divide by $x^2$:

Method: Reduce to standard linear form, then integrating factor

Use this for a linear equation that is NOT yet in standard form — for example one that starts as $x\dfrac{dy}{dx} + (\cdots)y = (\cdots)$. The coefficient of $\dfrac{dy}{dx}$ must be made $1$ before reading off $P(x)$.

Steps

Step 1: Divide through to make the coefficient of $\dfrac{dy}{dx}$ equal to $1$.

Only then is $\dfrac{dy}{dx} + P(x)y = Q(x)$ genuinely in standard form and $P(x)$ correct. Skipping this gives a wrong $P$ and a wrong integrating factor.

Step 2: Find the integrating factor $\mu(x) = e^{\int P\,dx}$.

Simplify exponentials of logs, e.g. $e^{2\log|x|} = x^2$.

Step 3: Multiply and recognise the exact derivative.

The left side becomes $\dfrac{d}{dx}\big(\mu y\big)$, so $\dfrac{d}{dx}(\mu y) = \mu Q$.

Step 4: Integrate and divide by $\mu$.

Common Mistakes

Mistake 1: Reading $P(x)$ before dividing by the leading coefficient.

Why it's wrong: the equation starts as $x\dfrac{dy}{dx} + 2y = x^2$; you must divide by $x$ to get $\dfrac{dy}{dx} + \dfrac{2}{x}y = x$ before $P(x) = \dfrac{2}{x}$ is correct. Using $P = 2$ gives a wrong integrating factor. Correct approach: make the coefficient of $\dfrac{dy}{dx}$ equal to $1$ first.

Mistake 2: Simplifying $e^{2\log|x|}$ incorrectly.

Why it's wrong: $e^{2\log|x|} = x^2$, not $2x$ or $2\log x$. Correct approach: use $e^{\log t} = t$ after writing $2\log|x| = \log(x^2)$.

Mistake 3: Forgetting to divide the constant term by $\mu$.

Why it's wrong: from $x^2 y = \dfrac{x^4}{4} + C$, dividing by $x^2$ gives $\dfrac{C}{x^2}$, not $C$. Correct approach: divide the whole right side, including $C$, by $x^2$.