## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 400: 3

#### Answer

See below the result.

#### Work Step by Step

Given: $y=(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt$ Use product rule. So, we get $y'=(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}$ and $x^2 y'=x^2[(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}]=-\int^{x}_1\dfrac{e^t}{t}dt+e^x$ or, $=-x[(\dfrac{1}{x})(\int^{x}_1\dfrac{e^t}{t}dt)]+(e^x)$ Thus, $x^2 y’=-xy+e^x$ Hence, this is our required differential equation.

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