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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 43–50, find the particular solution that satisfies the differential equation and the initial condition. See Example 6.

f ' ( x ) = x 2 5 x 2 , x > 0 ; f ( 1 ) = 2

To determine

To calculate: The particular solution of differential equation f'(x)=x25x2, x>0 with initial condition f(1)=2.

Explanation

Given Information:

The differential equation is f'(x)=x25x2, x>0 and the initial condition is f(1)=2.

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)=x25x2, 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=(x25x2)dxf(x)=x2x2dx5x2dx=1dx5x2dx

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

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