Chapter 6 - Differential Equations - 6.4 Exercises - Page 428: 19

$$y = \sin x + \left( {x + 1} \right)\cos x$$

$$\eqalign{ & y' + y\tan x = \sec x + \cos x,{\text{ }}y\left( 0 \right) = 1 \cr & {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} + y\tan x = \sec x + \cos x \cr & {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{With }}P\left( x \right) = \tan x{\text{ and }}Q\left( x \right) = \sec x + \cos x \cr & {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {\tan xdx} }} = {e^{\ln \left| {\sec x} \right|}} = \sec x \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & \sec x\left( {\frac{{dy}}{{dx}} + y\tan x} \right) = \sec x\left( {\sec x + \cos x} \right) \cr & \sec x\frac{{dy}}{{dx}} + y\sec x\tan x = {\sec ^2}x + 1 \cr & {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr & \frac{d}{{dx}}\left[ {y\sec x} \right] = {\sec ^2}x + 1 \cr & d\left[ {y\sec x} \right] = \left( {{{\sec }^2}x + 1} \right)dx \cr & {\text{Integrate both sides}} \cr & y\sec x = \int {\left( {{{\sec }^2}x + 1} \right)} dx \cr & y\sec x = \tan x + x + C \cr & {\text{Solve for }}y \cr & y = \frac{{\tan x}}{{\sec x}} + \frac{x}{{\sec x}} + \frac{C}{{\sec x}} \cr & y = \sin x + x\cos x + C\cos x{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}y\left( 0 \right) = 1 \cr & 1 = \sin \left( 0 \right) + 0\cos \left( 0 \right) + C\cos \left( 0 \right) \cr & 1 = C \cr & {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr & y = \sin x + x\cos x + \cos x \cr & y = \sin x + \left( {x + 1} \right)\cos x \cr} $$