## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 19

#### Answer

$$y =C e^{-\frac{5}{2} x}+\frac{4}{5}$$

#### Work Step by Step

Given $$2y'+5 y -4=0$$ \begin{aligned} 2 \frac{d y}{d x}+5 y &=4 \\ 2 \frac{d y}{d x}+5 y-5 y &=4-5 y \\ 2 \frac{d y}{d x} &=4-5 y \\ \frac{2}{4-5 y} \frac{d y}{d x} &=1\\ \int \frac{2}{4-5 y}dy&= \int dx\\ \frac{-2}{5}\ln |4-5y|&=x+C \end{aligned} Hence \begin{aligned} \ln (4-5 y) &=-\frac{5}{2} x-\frac{5}{2} c \\ y &=C e^{-\frac{5}{2} x}+\frac{4}{5} \end{aligned}

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