Q.Solve the following differential equation: $x \frac{dy}{dx} + 2y = x^2 \log x$

Concept: Linear First-Order ODE (Initial Value Problem not specified, so general solution).

The equation is

Step 1 — Standard form

Divide through by $x$ (for $x \neq 0$):

Step 2 — Integrating factor

Step 3 — Multiply and integrate

Multiplying:

Integrate both sides:

Using integration by parts ($u = \log x$, $dv = x^3 dx$): …

Why This Approach Works

The given equation is a first-order linear ordinary differential equation of the form:

The key idea: we multiply both sides by an integrating factor $\mu(x) = e^{\int P(x) dx}$, which turns the left-hand side into the exact derivative of $\mu(x) y$. This reduces the problem to a direct integration.

Let’s rewrite the equation in standard form first.

Step-by-Step Solution

1. Rewrite in standard linear form

We have:

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

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

2. Compute the integrating factor

We take $x^2$ (positive for $x>0$; the absolute value is handled by the constant later).

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

Multiply both sides of the standard form by $x^2$:

Notice the left side is exactly $\frac{d}{dx}(x^2 y)$ — check by differentiating:

So the equation becomes:

4. Integrate both sides

Integrate with respect to $x$:

Now evaluate the integral. Use integration by parts: let $u = \log x$, $dv = x^3 dx$. Then $du = \frac{1}{x} dx$, $v = \frac{x^4}{4}$.

$\int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} dx$ …

Method: Integrating factor with an $x^{n}\log x$ right-hand side

For a linear equation whose right side contains $\log x$, the integrating factor is a power of $x$ and the final integral is done by parts.

Steps

Step 1: Standard form.

Divide by the coefficient of $\dfrac{dy}{dx}$ to get $\dfrac{dy}{dx}+P\,y=Q$.

Step 2: Integrating factor from $P=\dfrac{n}{x}$.

Step 3: Integrate $\int x^{m}\log x\,dx$ by parts. …

Common Mistakes

Mistake 1: Reading $P$ from $xy'+2y$ without dividing by $x$.

Why it's wrong: standard form needs coefficient $1$ on $\dfrac{dy}{dx}$, giving $P=\tfrac2x$ and I.F. $=x^2$ — not $e^{2}$ or $e^{2x}$. Correct approach: divide by $x$ first.

Mistake 2: Integrating $\int x^3\log x\,dx$ incorrectly. …