Name: Linear Differential Equation: Illustration
Uploaded: 2019-07-19
Duration: 3 min 43 s
Description: You might be familiar with linear equations right from your middle school. Let's upgrade the concept to linear differential equations. This video teaches you...

Question

Find a particular solution of the differential equation $x \frac{d y}{d x} - y = \log x$ , given that $y = 0$ when $x = 1$ .

Accepted Answer

Given differential equation is $x \frac{d y}{d x} - y = \log x . . . . . . . . . . . . (1)$

Dividing above equation by $x$

$\frac{d y}{d x} - \frac{1}{x} y = \frac{\log x}{x} \dots \dots \dots (1)$

Equation $(1)$ is of the form $\frac{d y}{d x} + P y = Q$

Where, $P = \frac{- 1}{x}$ and $Q = \frac{\log x}{x}$

Therefore, integrating factor is $IF = e^{\int P d x}$

$= e^{\int \frac{- 1}{x} d x}$

$= e^{- \log (x)}$

$= e^{\log (\frac{1}{x})}$

$= \frac{1}{x}$

General solution is $y \cdot (lF) = \int Q \cdot (lF) d x + c$

$\Rightarrow y \cdot \frac{1}{x} = \int \frac{\log x}{x} \cdot \frac{1}{x} d x + c$

Let $I = \int (\frac{\log x}{x}) \cdot (\frac{1}{x}) d x$

Put $\log x = t$

$(\frac{1}{x}) d x = d t$

$I = \int (\frac{t}{e^{t}}) d t$

Therefore $I = \int t \cdot e^{- t} d t$

Let $u = t$ and $v = e^{- t}$

$\Rightarrow I = t \cdot \int e^{- t} d t - \int (\frac{d}{d t} (t) \cdot \int e^{- t} d t) d t$

$\Rightarrow I = - t . e^{- t} - \int ((1) \cdot (- e^{- t})) d t$

$\Rightarrow I = - t \cdot e^{- t} - e^{- t}$

$\Rightarrow I = - \frac{\log x}{x} - \frac{1}{x}$

Substituting $I$ in $(2),$ we get

$y \cdot (\frac{1}{x}) = - \frac{\log x}{x} - \frac{1}{x} + c$

$\Rightarrow y = - \log x - 1 + c x$

For particular solution put $y = 0$ and $x = 1$ in above equation, we get $c = 1$

Substituting $c$ in general solution, we get particular solution as $y = x - \log x - 1$ .

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