(vi) The solution of the differential equation $x\frac{dy}{dx}+2y=x^2$ is ______.

> [!TLDR] > This is a first-order linear differential equation solved using the integrating factor method. The general solution is $y = \frac{x^2}{4} + \frac{C}{x^2}$. The key insight: when you see a differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor method is your go-to tool. Here, the equation is $x\frac{dy}{dx} + 2y = x^2$, which needs a small rearrangement first. Let’s work through it step by step. 1. **Rewrite in standard form** Divide through by $x$ (assuming $x \neq 0$): $$\frac{dy}{dx} + \frac{2}{x}y = x$$ Now it matches $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \frac{2}{x}$ and $Q(x) = x$. 2. **Find the integrating factor** The integrating factor (IF) is $e^{\int P(x)\,dx}$. $$\int P(x)\,dx = \int \frac{2}{x}\,dx = 2\log|x| = \log(x^2)$$ So the IF is $e^{\log(x^2)} = x^2$. > [!TIP] > Remember: $e^{\log(\text{something})} = \text{something}$, so no need to overcomplicate — the IF is simply $x^2$. 3. **Multiply through by the IF** Multiply the standard-form equation by $x^2$: $$x^2\frac{dy}{dx} + 2x y = x^3$$ Notice the left side is now the derivative of $y \cdot \text{IF}$: $$\frac{d}{dx}(x^2 y) = x^3$$ …

(vi) The solution of the differential equation x(dy)/(dx)+2y=x² is ___.

NCERT Exemplar · Q39

x\frac{dy}{dx}+2y=x^2

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Concept understanding — Integrating Factor Method

Integrating Factor Method

Some first-order differential equations refuse to separate — you cannot get all the $y$ 's on one side and all the $x$ 's on the other. The integrating factor method is the standard trick for a special (and very common) family of these: the linear first-order equation. The idea is beautifully simple: multiply the whole equation by one cleverly chosen function, and the messy left-hand side collapses into a single derivative that we can integrate directly.

The Standard Form

An equation is linear of first order if it can be written as

\frac{dy}{dx} + P\,y = Q

where $P$ and $Q$ are functions of $x$ alone (or constants). Notice $y$ and $\frac{dy}{dx}$ appear only to the first power, and never multiplied together — that is what "linear" means here.

Tip

Always rearrange into this exact shape first. The coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$ and $Q$ .

The Integrating Factor

The magic multiplier is

\text{I.F.} = e^{\int P\,dx}.

Why this one? Multiply the equation by $e^{\int P\,dx}$ :

e^{\int P\,dx}\frac{dy}{dx} + P\,e^{\int P\,dx}\,y = Q\,e^{\int P\,dx}.

By the product rule, the left-hand side is exactly $\dfrac{d}{dx}\!\left(y\cdot e^{\int P\,dx}\right)$ , because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$ . So the equation becomes

\frac{d}{dx}\big(y\cdot \text{I.F.}\big) = Q\cdot \text{I.F.}

The left side is now a single derivative — that is the whole point of choosing this factor.

The Solution

Integrate both sides with respect to $x$ :

y\cdot \text{I.F.} = \int \big(Q\cdot \text{I.F.}\big)\,dx + C.

This is the general solution. In words: (solution) × (integrating factor) = integral of (Q × integrating factor), plus a constant.

A Quick Illustration

For $\dfrac{dy}{dx} + \dfrac{1}{x}\,y = x$ , we read $P = \tfrac1x$ , $Q = x$ . Then $\int P\,dx = \log x$ , so $\text{I.F.} = e^{\log x} = x$ . The solution is

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

--- …

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This is a first-order linear differential equation solved using the integrating factor method. The general solution is $y = \frac{x^2}{4} + \frac{C}{x^2}$ .

The key insight: when you see a differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ , the integrating factor method is your go-to tool. Here, the equation is $x\frac{dy}{dx} + 2y = x^2$ , which needs a small rearrangement first.

Let’s work through it step by step.

Rewrite in standard form Divide through by $x$ (assuming $x \neq 0$ ):

\frac{dy}{dx} + \frac{2}{x}y = x

Now it matches $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \frac{2}{x}$ and $Q(x) = x$ .

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

\int P(x)\,dx = \int \frac{2}{x}\,dx = 2\log|x| = \log(x^2)

So the IF is $e^{\log(x^2)} = x^2$ .

Tip

Remember: $e^{\log(\text{something})} = \text{something}$ , so no need to overcomplicate — the IF is simply $x^2$ .

Multiply through by the IF Multiply the standard-form equation by $x^2$ :

x^2\frac{dy}{dx} + 2x y = x^3

Notice the left side is now the derivative of $y \cdot \text{IF}$ :

$\frac{d}{dx}(x^2 y) = x^3$ …

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Q.(vi) The solution of the differential equation $x\frac{dy}{dx}+2y=x^2$ is ______.

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Integrating Factor for a Linear Equation Not Yet in Standard Form

Steps

Common Mistakes

Q.(vi) The solution of the differential equation xdydx+2y=x2x\frac{dy}{dx}+2y=x^2xdxdy​+2y=x2 is ______.

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Q.(vi) The solution of the differential equation $x\frac{dy}{dx}+2y=x^2$ is ______.

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