CK-12-Calculus

http://www.ck12.org Chapter 4. Integration

Solution:
We can re-state the problem in terms of a differential equation that satisfies an initial condition.
f′(x) = 3 x+4 withf( 2 ) =6.
By integrating the right side of the differential equation we have

f(x) =^32 x^2 + 4 x+Cas the general solution. Substituting the condition thatf( 2 ) =6 gives

6 =^32 ( 2 )^2 + 4 ( 2 )+C,

6 = 6 + 8 +C,

C=− 8.

Hencef(x) =^32 x^2 + 4 x−8 is theparticular solutionof the original equation f′(x) = 3 x+4 satisfying theinitial
conditionf( 2 ) = 6.
Finally, since we are interested in the valuef(− 2 ), we put−2 into our expression forfand obtain:

f(− 2 ) =− 10

Lesson Summary

1. We found general solutions of differential equations.


2. We used initial conditions to find particular solutions of differential equations.

Multimedia Link

The following applet allows you to set the initial equation forf′(x)and then the slope field for that equation is
displayed. In magenta you’ll see one possible solution forf(x). If you move the magenta point to the initial value,
then you will see the graph of the solution to the initial value problem. Follow the directions on the page with the
applet to explore this idea, and then try redoing the examples from this section on the applet. Slope Fields Applet.

Review Questions

In problems #1–3, solve the differential equation forf(x).

1. f′(x) = 2 e^2 x− 2 √x


2. f′(x) =sinx−e^1 x


3. f′′(x) = ( 2 +x)√x

In problems #4–7, solve the differential equation forf(x)given the initial condition.

4. f′(x) = 6 x^5 − 4 x^2 +^73 andf( 1 ) = 4