   Chapter 6.3, Problem 21E

Chapter
Section
Textbook Problem

Finding a General Solution Using Separation of Variables In Exercises 19-28, find the particular solution of the differential equation that satisfies the initial condition.Differential Equation Initial Condition y ( x + 1 ) + y ′ = 0 y ( − 2 ) = 1

To determine

To calculate: The particular solution of the differential equation given as, y(x+1)+y'=0 with the initial condition, y(2)=1.

Explanation

Given:

The differential equation is y(x+1)+y'=0 with the initial condition y(2)=1.

Formula used:

The differential equation is in the variable separable form, f(y)dy=g(x)dx where f(y) and g(x) are functions of y and x respectively.

The formula of integration is xndx=xn+1n+1+C.

Calculation:

Consider the differential equation given as,

y(x+1)+y'=0

Simplify,

dydx=y(x+1)dyy=(x+1)dx

This differential equation is of the variable separable form, f(y)dy=g(x)dx

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