   Chapter 12.5, Problem 32E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# In Problems 29-36, find the particular solution to each differential equation. 34.   ( x + 1 )   d y = y 2   d x  when  x = 0 ,   y = 2

To determine

To calculate: The particular solution to the differential equation (x+1)dy=y2dx when x=0, y=2.

Explanation

Given Information:

The provided differential equation is (x+1)dy=y2dx and the values are x=0, y=2.

Formula used:

Solution of the differential equation g(y)dy=f(x)dx is g(y)dy=f(x)dx.

The power of x formula of integration is xndx=xn+1n+1+C, where n1.

The logarithmic formula of integration is 1x+adx=ln|x+a|+C, where a is any real number.

Calculation:

Consider the differential equation, (x+1)dy=y2dx

Rearrange the equation as,

1y2dy=1x+1dxy2dy=1x+1dx

Integrate both sides to get:

y

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