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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