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CalculusCalculus: An Applied Approach (MindTap Course List)10th Edition

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Chapter 5.1, Problem 52E
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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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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 ,   f ' ( 0 ) = 6 ,   f ( 0 ) = 3

To determine

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

Explanation

Given Information:

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

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

∫f''(x)dx=∫x2dxf'(x)=[x2+12+1]+C1=x33+C1

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

Substitute 0 for x in differential equation f'(x)=x33+C1.

f'(0)=(0)33+C1=C1

Substitute 6 for f'(0) in above differential equation to get the value of constant C1.

6=C1

Substitute 6 for C1 in the function f'(x)=x33+C1.

f'(x)=x33+6

Integrate the above differential equation f'(x)=x33+6, use the simple power rule of integration ∫xndx=xn+1n+1+C

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