   Chapter 6.3, Problem 24E

Chapter
Section
Textbook Problem

Finding a Particular 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 1 = x 2 y ' − x 1 − y 2 = 0 y ( 0 ) = 1

To determine

To calculate: The particular solution of the differential function using separation of variables satisfying the initial condition, y1x2y'x1y2=0 for y(0)=1.

Explanation

Given:

The differential function y1x2y'x1y2=0 for y(0)=1.

Formula used:

xndx=xn+1n+1n1

Calculation:

For the given function, y1x2y'x1y2=0 for y(0)=1, the variables are separated calculated as follow,

y1x2y'x1y2=0y1x2dy=x1y2dxy1y2dy=x1x2dx

Now, integrate normally for, y and x,

y1y2dy=x1x2dx

Let, 1y2=u2ydy=du

Similarly, 1x2</

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