   Chapter 6.4, Problem 20E

Chapter
Section
Textbook Problem

Finding a Particular Solution In Exercises 17-24, find the particular solution of the first-order linear differential equation for x > 0 that satisfies the initial condition.Differential Equation Initial Condition y ' + y sec x = sec x y ( 0 ) = 4

To determine

To calculate: The particular solution of the given first order linear differential equation

y+ysecx=secx.

Explanation

Given:

The given differential equation is expressed as y+ysecx=secx,

With the initial conditions at y(0)=4.

Formula used:

The general solution of linear differential equation dydx+P(x)y=Q(x) is given by,

y(x)=eP(x)dxQ(x)+ceP(x)dx

Here, the term eP(x)dx is known as the integrating factor.

Calculation:

The given differential equation is expressed as y+ysecx=secx.

From the equation above,

P(x)=secxQ(x)=secx

Therefore, integrating factor is,

eP(x)dx=e(secx)dx=eln|tanx+secx|=tanx+secx

So, the general solution can be written as,

y(x)=(tanx+secx)(secx)dx+ctanx+

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