   Chapter 5.1, Problem 53E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding a Particular Solution In Exercises 51–54, find a function f that satisfies the differential equation and the initial conditions. f " ( x ) = x − 2 / 3 ,   f ' ( 8 ) = 6 ,   f ( 0 ) = 0

To determine

To calculate: The particular solution of differential equation f''(x)=x2/3 with initial condition f'(8)=6 and f(0)=0.

Explanation

Given Information:

The differential equation is f''(x)=x2/3, and the initial condition are f'(8)=6 and f(0)=3.

Formula used:

The simple power rule of integration xndx=xn+1n+1+C.

Calculation:

Consider the differential equation, f''(x)=x2/3.

Integrate the provided differential equation, use the constant rule of integration xndx=xn+1n+1+C.

f''(x)dx=x2/3dxf'(x)=[x23+123+1]+C1=x1313+C1=3x13+C1

Now, from the initial condition of the differential function f'(x).

Substitute 8 for x in differential equation f'(x)=3x13+C1.

f'(8)=3(8)13+C1=3(23)13+C1=3(2)+C1=6+C1

Substitute 6 for f'(8) in above differential equation to get the value of constant C1.

6=6+C1C1=66=0

Substitute 0 for C1 in the differential function f'(x)=3x13+C1

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