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Tuesday, 5 August 2014

CHAPTER 15 - Miscellaneous Examples

     Example:  1   

Let f:{\mathbb{R}^ + } \to \mathbb{R}  be a differentiable function such that
f\left( x \right) = e - \left( {x - 1} \right)\ln \dfrac{x}{e} + \int\limits_1^x {f\left( x \right)dx}
Find a simple expression for f(x).
Solution:  1

Step-1

Differentiating the given relation, we have
\dfrac{{df}}{{dx}} =  - \dfrac{{\left( {x - 1} \right)}}{x} - \ln \dfrac{x}{e} + f
 \Rightarrow  \dfrac{{df}}{{dx}} - f = \dfrac{1}{x} - \ln x

Step-2

This is evidently a first-order linear DE; the IF is {e^{\int { - dx} }} = {e^{ - x}}. Multiplying it across both sides of the DE renders the DE exact and its solution is given by
{e^{ - x}} \cdot f = \int {{e^{ - x}}\left( {\dfrac{1}{x} - \ln x} \right)dx}
 = {e^{ - x}}\ln x + C
 \Rightarrow  f\left( x \right) = \,\ln x + C\,{e^x}\ldots (1)

Step-3

From the relation specified in the equation, note that
f\left( 1 \right) = e - \left( {1 - 1} \right)\left( {\ln \dfrac{1}{e}} \right) + \int\limits_1^1 {f\left( x \right)dx}
=e
From (1) f(1) = Ce. This gives C = 1. Thus, the function f(x) has the simple form
f\left( x \right) = \ln x + {e^x}
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