   Chapter 5, Problem 16RE ### 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 15–18, find the particular solution that satisfies the differential equation and the initial condition. f ' ( x ) = 3 x 2 − 8 x ;   f ( 1 ) = 12

To determine

To calculate: The particular solution that satisfied the differential equation and initial condition f(x)=3x28x and  f(1)=12.

Explanation

Given Information:

The provided differential equation f(x)=3x28x.

The initial condition f(1)=12

Formula used:

The power rule of integrals:

undu=un+1n+1+C (for n1)

Here, u is function of x.

The property of Intro-differential:

df(x)dxdx=f(x)

Calculation:

Consider the derivative equation:

f(x)=3x28x

Rewrite the integrand as:

df(x)dx=3x28x

Apply, integration on both sides:

df(x)dxdx=(3x28x)dx+C

Now apply the power rule of integration and property of Intro-differential:

f(x)=(

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