   Chapter 12.5, Problem 35E ### 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. 35.  2 x y d y d x = y 2 + 1  when  x = 1 ,   y = 2

To determine

To calculate: The particular solution to the differential equation 2xydydx=y2+1 when x=1,y=2.

Explanation

Given Information:

The provided differential equation is 2xydydx=y2+1 and the values are x=1,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 1xdx=ln|x|+C.

The product rule for any positive numbers M and N, ln(MN)=lnM+lnN.

Calculation:

Consider the differential equation, 2xydydx=y2+1

Rearrange the equation as,

2yy2+1dy=1xdx

Integrate both sides of the equation,

2yy2+1dy=1xdx

If y2+1=u then,

2ydy=du

Substitute y2+1=u and 2ydy=du in the equation 2yy2+1dy=1xdx

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