Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.6 First-Order Linear Differential Equations - Problems - Page 59: 12

Answer

\[y(x)=\sin x+C\cos x\]

Work Step by Step

$(1-y\sin x)dx-(\cos x)dy=0$ ___(1) $(1-y\sin x)dx=(\cos x)dy$ $\Large\frac{dy}{dx}$= $\sec x-y\tan x$ $\Large\frac{dy}{dx}$ +$(\tan x)y=\sec x$ This is linear differential equation Integrating factor:- $I(x)=e^{\int\tan x\;dx}=e^{\ln|\sec x|}=\sec x$ Multiply (1) by $\sec x$ $\sec x\Large\frac{dy}{dx}$ +$\sec x\tan x\; y=\sec^2 x$ $\Large\frac{d}{dx}$ $[y\sec x]=\sec ^2 x$ Integrating, $y\sec x=\int\sec^2 x\;dx+C$ $C$ is constant of integration $y\sec x=\tan x +C$ $y=\sin x+C\cos x$ Hence general solution of (1) is $y(x)=\sin x+C\cos x$.
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