# Solve the Following Initial Value Problem: X D Y D X − Y = Log X , Y ( 1 ) = 0 - Mathematics

Sum

Solve the following initial value problem:-
$x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0$

#### Solution

We have,
$x\frac{dy}{dx} - y = \log x$
$\Rightarrow \frac{dy}{dx} - \frac{y}{x} = \frac{\log x}{x} . . . . . \left( 1 \right)$
Clearly, it is a linear differential equation of the form
$\frac{dy}{dx} + Py = Q$
$\text{ where }P = - \frac{1}{x}\text{ and }Q = \frac{\log x}{x}$
$\therefore I.F. = e^{\int P\ dx}$
$= e^{- \int\frac{1}{x} dx}$
$= e^{- \log x}$
$= \frac{1}{x}$
$\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = \frac{1}{x}, \text{ we get }$
$\frac{1}{x} \left( \frac{dy}{dx} - \frac{1}{x}y \right) = \frac{1}{x} \times \frac{\log x}{x}$
$\Rightarrow \frac{1}{x}\frac{dy}{dx} - \frac{1}{x^2}y = \frac{\log x}{x^2}$
Integrating both sides with respect to x, we get

$\Rightarrow \frac{y}{x} = \log x\int\frac{1}{x^2}dx - \int\left[ \frac{d}{dx}\left( \log x \right)\int\frac{1}{x^2}dx \right]dx + C$
$\Rightarrow \frac{y}{x} = - \frac{\log x}{x} + \int\frac{1}{x^2}dx + C$
$\Rightarrow \frac{y}{x} = - \frac{\log x}{x} - \frac{1}{x} + C$
$\Rightarrow y = - \log x - 1 + Cx . . . . . . . . . \left( 2 \right)$
Now,
$y\left( 1 \right) = 0$
$\therefore 0 = - 0 - 1 + C\left( 1 \right)$
$\Rightarrow C = 1$
$\text{ Putting the value of C in }\left( 2 \right),\text{ we get }$
$y = - \log x - 1 + x$
$\Rightarrow y = x - 1 - \log x$
$\text{ Hence, }y = x - 1 - \log x\text{ is the required solution .}$

Concept: Basic Concepts of Differential Equation
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Exercise 22.1 | Q 37.02 | Page 107