(vii) The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ______.

> [!TLDR] > This is a first-order linear differential equation. We rewrite it in standard form, find the integrating factor $1+x^2$, and integrate to get the general solution: $y(1+x^2) = \frac{4x^3}{3} + C$. The key to solving any first-order linear differential equation is recognising its standard form and using an integrating factor. Let's see why that works here. The given equation is $(1+x^2)\frac{dy}{dx} + 2xy - 4x^2 = 0$. This is linear in $y$ because $y$ and $\frac{dy}{dx}$ appear only to the first power, and there's no product like $y\frac{dy}{dx}$. The standard form for such an equation is $\frac{dy}{dx} + P(x)y = Q(x)$, where the coefficient of $\frac{dy}{dx}$ is 1. Once we have that, the integrating factor $\mu(x) = e^{\int P\,dx}$ makes the left side a perfect derivative, so we can integrate both sides directly. Let's work through it. 1. **Rewrite in standard form.** Divide every term by $1+x^2$ (which is never zero for real $x$, so no worries about losing solutions): $$ \frac{dy}{dx} + \frac{2x}{1+x^2}\,y = \frac{4x^2}{1+x^2}. $$ Here $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$. 2. **Find the integrating factor.** $$ \mu(x) = e^{\int P\,dx} = e^{\int \frac{2x}{1+x^2}\,dx}. $$ The integral $\int \frac{2x}{1+x^2}\,dx$ is a standard logarithmic form: let $u = 1+x^2$, then $du = 2x\,dx$, so the integral becomes $\int \frac{du}{u} = \log|1+x^2|$. Since $1+x^2 > 0$, we drop the absolute value: $$ \int \frac{2x}{1+x^2}\,dx = \log(1+x^2). $$ Hence, $$ \mu(x) = e^{\log(1+x^2)} = 1+x^2. $$ > [!TIP] > Notice that the integrating factor turned out to be exactly the original coefficient of $\frac{dy}{dx}$. This often happens when $P(x)$ is of the form $\frac{f'(x)}{f(x)}$ — a neat shortcut to spot. 3. **Multiply the standard-form equation by $\mu(x)$.** $$ (1+x^2)\frac{dy}{dx} + 2x\,y = 4x^2. $$ The left side is now the derivative of $\mu(x) \cdot y$: $$ \frac{d}{dx}\

(vii) The solution of (1+x²)(dy)/(dx)+2xy-4x²=0 is ___.

NCERT Exemplar · Q40

(1+x^2)\frac{dy}{dx}+2xy-4x^2=0

Uttar Pradesh UpmspShort· 1mImportance★★★★★

72% · 160/222 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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.

--- …

Start your 14-day free trial to unlock the full solution →

This is a first-order linear differential equation. We rewrite it in standard form, find the integrating factor $1+x^2$ , and integrate to get the general solution: $y(1+x^2) = \frac{4x^3}{3} + C$ .

The key to solving any first-order linear differential equation is recognising its standard form and using an integrating factor. Let's see why that works here.

The given equation is $(1+x^2)\frac{dy}{dx} + 2xy - 4x^2 = 0$ . This is linear in $y$ because $y$ and $\frac{dy}{dx}$ appear only to the first power, and there's no product like $y\frac{dy}{dx}$ . The standard form for such an equation is $\frac{dy}{dx} + P(x)y = Q(x)$ , where the coefficient of $\frac{dy}{dx}$ is 1. Once we have that, the integrating factor $\mu(x) = e^{\int P\,dx}$ makes the left side a perfect derivative, so we can integrate both sides directly.

Let's work through it.

Rewrite in standard form. Divide every term by $1+x^2$ (which is never zero for real $x$ , so no worries about losing solutions):

\frac{dy}{dx} + \frac{2x}{1+x^2}\,y = \frac{4x^2}{1+x^2}.

Here $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$ .

Find the integrating factor.

\mu(x) = e^{\int P\,dx} = e^{\int \frac{2x}{1+x^2}\,dx}.

The integral $\int \frac{2x}{1+x^2}\,dx$ is a standard logarithmic form: let $u = 1+x^2$ , then $du = 2x\,dx$ , so the integral becomes $\int \frac{du}{u} = \log|1+x^2|$ . Since $1+x^2 > 0$ , we drop the absolute value:

\int \frac{2x}{1+x^2}\,dx = \log(1+x^2).

Hence,

\mu(x) = e^{\log(1+x^2)} = 1+x^2.

Tip

Notice that the integrating factor turned out to be exactly the original coefficient of $\frac{dy}{dx}$ . This often happens when $P(x)$ is of the form $\frac{f'(x)}{f(x)}$ — a neat shortcut to spot.

Multiply the standard-form equation by $\mu(x)$ .

(1+x^2)\frac{dy}{dx} + 2x\,y = 4x^2.

The left side is now the derivative of $\mu(x) \cdot y$ :

$\frac{d}{dx}\big[(1+x^2)y\big] = 4x^2.$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ39 — (vi) The solution of the differential equation x(dy)/(dx)+2y=x² is ___…→Next questionQ41 — (viii) The solution of the differential equation ydx+(x+xy)dy=0 is ___…

Q.(vii) The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ______.

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Recognising the Coefficient Itself as the Integrating Factor

Steps

Common Mistakes

Q.(vii) The solution of (1+x2)dydx+2xy−4x2=0(1+x^2)\frac{dy}{dx}+2xy-4x^2=0(1+x2)dxdy​+2xy−4x2=0 is ______.

🎓Unlock everything free for 14 days

Q.(vii) The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ______.

Unlock everything free for 14 days