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

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Chapter 5.1, Problem 49E
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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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BuyFindarrow_forward

Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Finding a Particular Solution In Exercises 43–50, find the particular solution that satisfies the differential equation and the initial condition. See Example 6.

f ' ( x ) = 2 − x x 3 ,   x > 0 ;   f ( 2 ) = 3 4

To determine

To calculate: The particular solution of differential equation f'(x)=2−xx3, x>0 with initial condition f(2)=34.

Explanation

Given Information:

The differential equation is f'(x)=2−xx3, x>0 and the initial condition is f(2)=34.

Formula used:

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

The sum rule of the integration ∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx.

Calculation:

Consider the differential equation, f'(x)=2−xx3, x>0.

Integrate the provided differential equation, use the sum rule of the integration ∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx.

∫f'(x)dx=∫(2−xx3)dxf(x)=∫2x3dx−∫xx3dx=2∫x−3dx−∫x−2dx

Integrate the further, use the simple power rule of integration ∫xndx=xn+1n+1+C

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