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

Concept: Linear First-Order ODE (Integrating Factor Method)

This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \frac{1}{x}$ and $Q(x) = x^2$.

Step 1: Find the integrating factor

$\mu(x) = e^{\int P(x)\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\log x} = x$.

Step 2: Multiply through and rewrite

$x\frac{dy}{dx} + y = x^3 \quad\Rightarrow\quad \frac{d}{dx}(x y) = x^3$.

Step 3: Integrate both sides

$x y = \int x^3\,dx = \frac{x^4}{4} + C$.

Step 4: Solve for $y$

$y = \frac{x^3}{4} + \frac{C}{x}$.

We start with the equation:

This is a classic first-order linear ordinary differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$. The key idea is to multiply both sides by an integrating factor that turns the left-hand side into the derivative of a product. This works because the left side already looks like a product rule waiting to happen — we just need the right multiplier.

Here, $P(x) = \frac{1}{x}$ and $Q(x) = x^2$.

1. Find the integrating factor The integrating factor $\mu(x)$ is given by:

For $x > 0$ (which is typical unless otherwise stated), we take $\mu(x) = x$.

2. Multiply the entire equation by $\mu(x)$

Simplifies to:

3. Recognize the left side as a derivative Notice that:

So the equation becomes:

4. Integrate both sides

where $C$ is the constant of integration.

5. Solve for $y$ Divide through by $x$ (assuming $x \neq 0$):

Method: Integrating factor when $P=\dfrac{1}{x}$ gives I.F. $=x$

This is the standard route for a linear equation whose coefficient is a power of $x$, where the integrating factor comes out as a power of $x$.

Steps

Step 1: Standard form.

Ensure the coefficient of $\dfrac{dy}{dx}$ is $1$ and read off $P(x)$ and $Q(x)$.

Step 2: Integrate $P$ and exponentiate.

For $P=\dfrac{n}{x}$, $\int P\,dx=n\log x$, so

Step 3: Multiply and recognise the derivative.

The left side becomes $\dfrac{d}{dx}\big(x^{n}y\big)$. Integrate the right side:

Step 4: Solve for $y$.

Divide by $x^{n}$; the constant term becomes $\dfrac{C}{x^{n}}$, valid for $x\neq0$.

Common Mistakes

Mistake 1: Getting the integrating factor wrong from $P=\tfrac1x$.

Why it's wrong: $\int\tfrac1x\,dx=\log x$, so I.F. $=e^{\log x}=x$, not $e^{x}$ or $\tfrac1x$. Correct approach: exponentiate the log to a power of $x$.

Mistake 2: Not multiplying $Q$ by the I.F. before integrating.

Why it's wrong: after multiplying, the right side is $x\cdot x^2=x^3$, so $xy=\int x^3\,dx=\tfrac{x^4}{4}+C$. Integrating $x^2$ instead gives a wrong power. Correct approach: integrate $Q\cdot\text{I.F.}=x^3$.

Mistake 3: Dropping the $\tfrac{C}{x}$ term.

Why it's wrong: dividing $xy=\tfrac{x^4}{4}+C$ by $x$ leaves $\tfrac{C}{x}$, valid for $x\neq0$. Correct approach: keep it as part of the general solution.