   Chapter 12.5, Problem 33E ### 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 2 e 2 y   d x = ( x 3 + 1 )   d x  when  x = 1 ,   y = 0

To determine

To calculate: The particular solution to the differential equation x2e2ydy=(x3+1)dx when x=1,y=0.

Explanation

Given Information:

The provided differential equation is x2e2ydy=(x3+1)dx and the values are x=1,y=0.

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 exponential formula of integration is eaxdx=1aeax+C, where a is any real number.

Calculation:

Consider the differential equation, x2e2ydy=(x3+1)dx

Rearrange the equation as,

e2ydy=x3+1x2dxe2ydy=(x+x2)dx

Integrate both sides of the equation,

e2ydy=(x+x2)dx12e2y=x1+11+1+x2+12+1+C12e2y=x221x+C

The provided values are x=1,y=0

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