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

Solve the following differential equation: x log x (dy)/(dx) + y = (2)/(x) log x

Exercise 9.5 · Q7

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

Chhattisgarh CgbseTextbookSubjective· 3mImportance★★★★★

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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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The general solution is $y=\dfrac{Cx-2(\log x+1)}{x\log x}$ .

This is a first-order linear ODE. Divide by $x\log x$ to reach standard form:

\frac{dy}{dx}+\frac{1}{x\log x}\,y=\frac{2}{x^2},\qquad P(x)=\frac{1}{x\log x},\; Q(x)=\frac{2}{x^2}.

Integrating factor. With $u=\log x$ , $\displaystyle\int\frac{dx}{x\log x}=\log|\log x|$ , so $\mu(x)=e^{\log|\log x|}=\log x$ .

Multiply through. The left side becomes an exact derivative:

\frac{d}{dx}\big(y\log x\big)=\frac{2\log x}{x^2}.

Integrate the right side by parts ( $u=\log x,\; dv=\tfrac{2}{x^2}dx,\; v=-\tfrac{2}{x}$ ):

$\int\frac{2\log x}{x^2}\,dx=-\frac{2\log x}{x}+2\int\frac{dx}{x^2}=-\frac{2\log x}{x}-\frac{2}{x}+C.$ …

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

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Integrating factor when $\int P\,dx=\log(\log x)$

Steps

Common Mistakes

Q.Solve the following differential equation: xlog⁡xdydx+y=2xlog⁡xx \log x \frac{dy}{dx} + y = \frac{2}{x} \log xxlogxdxdy​+y=x2​logx

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

Unlock everything free for 14 days