Chapter 6.3, Problem 24E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
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 y1âˆ’x2y'âˆ’x1âˆ’y2=0 for y(0)=1.

Formula used:

âˆ«xndx=xn+1n+1,Â nâ‰ âˆ’1Â

Calculation:

For the given function, y1âˆ’x2y'âˆ’x1âˆ’y2=0 for y(0)=1, the variables are separated calculated as follow,

y1âˆ’x2y'âˆ’x1âˆ’y2=0y1âˆ’x2dy=x1âˆ’y2dxy1âˆ’y2dy=x1âˆ’x2dx

Now, integrate normally for, yÂ andÂ x,

y1âˆ’y2dy=x1âˆ’x2dx

Let, 1âˆ’y2=uâ‡’âˆ’2ydy=du

Similarly, 1âˆ’x2</

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