Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 19


$$y(x)=\frac{1}{5} e^{2 x}-\frac{6}{5} e^{-3 x}$$

Work Step by Step

Given $$y^{\prime}+3 y=e^{2 x}, \quad y(0)=-1$$ This a linear equation with $p(x) =3$ and $q(x) =e^{2x}$; then \begin{align*} \mu(x) &= e^{\int p(x)dx}\\ &=e^{\int3 dx}\\ &= e^{3x} \end{align*} Hence, the general solution is \begin{align*} \mu(x) y &=\int \mu (x)xq(x)dx\\ e^{3x} y&=\int e^{3x} e^{2x}dx \\ &=\int e^{5x}dx\\ &=\frac{1}{5}e^{5x}+C \end{align*} Then $$ y= \frac{1}{5}e^{2x}+C e^{-3x}$$ Since $ y(0)= -1 $, then $C= \frac{-6}{5}$, hence $$y(x)=\frac{1}{5} e^{2 x}-\frac{6}{5} e^{-3 x}$$
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