<meta http-equiv="refresh" content="1; url=/nojavascript/"> The Initial Value Problem | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Calculus Go to the latest version.

# 4.2: The Initial Value Problem

Created by: CK-12

## Learning Objectives

• Find general solutions of differential equations
• Use initial conditions to find particular solutions of differential equations

## Introduction

In the Lesson on Indefinite Integrals Calculus we discussed how finding antiderivatives can be thought of as finding solutions to differential equations: $F'(x) = f(x).$ We now look to extend this discussion by looking at how we can designate and find particular solutions to differential equations.

Let’s recall that a general differential equation will have an infinite number of solutions. We will look at one such equation and see how we can impose conditions that will specify exactly one particular solution.

Example 1:

Suppose we wish to solve the following equation:

$f'(x) = e^{3x} - 6\sqrt{x}.$

Solution:

We can solve the equation by integration and we have

$f(x) =\frac{1} {3} e^{3x} - 4x^{\frac{3} {2}} + C.$

We note that there are an infinite number of solutions. In some applications, we would like to designate exactly one solution. In order to do so, we need to impose a condition on the function $f.$ We can do this by specifying the value of $f$ for a particular value of $x.$ In this problem, suppose that we add the condition that $f(0) = 1.$ This will specify exactly one value of $C$ and thus one particular solution of the original equation:

Substituting $f (0) = 1$ into our general solution $f(x) = \frac{1} {3} e^{3x} - 4x^{\frac{3} {2}} + C$ gives $1 = \frac{1} {3} e^{3(0)} - 4(0)^{\frac{3} {2}} + C$ or $C = 1 - \frac{1} {3} = \frac{2} {3}.$ Hence the solution $f(x) = \frac{1} {3} e^{3x} - 4x^{\frac{3} {2}} + \frac{2} {3}$ is the particular solution of the original equation $f'(x) = e^{3x} - 6\sqrt{x}$ satisfying the initial condition $f (0) = 1.$

We now can think of other problems that can be stated as differential equations with initial conditions. Consider the following example.

Example 2:

Suppose the graph of $f$ includes the point $(2 ,6)$ and that the slope of the tangent line to $f$ at any point $x$ is given by the expression $3 x + 4.$ Find $f (-2).$

Solution:

We can re-state the problem in terms of a differential equation that satisfies an initial condition.

$f'(x) = 3 x + 4$ with $f (2) = 6$.

By integrating the right side of the differential equation we have

$f(x) & = \frac{3} {2} x^2 + 4x +C\ \text{as the general solution.\ Substituting the condition that} f (2) = 6\ \text{gives} \\ 6 & = \frac{3} {2} (2)^2 + 4(2) + C, \\6 & = 6 + 8 + C, \\C & = -8.$

Hence $f(x) = \frac{3} {2} x^2 + 4x - 8$ is the particular solution of the original equation $f'(x) = 3 x + 4$ satisfying the initial condition $f (2) = 6.$

Finally, since we are interested in the value $f (-2)$, we put $-2$ into our expression for $f$ and 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.

The following applet allows you to set the initial equation for $f'(x)$ and then the slope field for that equation is displayed. In magenta you'll see one possible solution for $f(x)$. 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 for $f (x).$

1. $f'(x) = 2e^{2x} - 2\sqrt{x}$
2. $f'(x) = \sin x - \frac{1} {e^x}$
3. $f''(x) = (2 + x) \sqrt{x}$

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

1. $f'(x) = 6x^5 - 4x^2 + \frac{7} {3}$ and $f(1) = 4$
2. $f'(x) = 3 x^2 + e^{2 x}$ and $f (0) = 3.$
3. $f'(x) = \sqrt[3]{x^2} - \frac{1} {x^2}$ and $f(1) = 3$
4. $f'(x) = (2\cos x - \sin x),$ $- \frac{\pi} {2} \le x \le \frac{\pi} {2},$ and $f\left ( \frac{\pi} {3}\right ) = \sqrt{3} + \frac{1} {2}$
5. Suppose the graph of $f$ includes the point $(-2, 4)$ and that the slope of the tangent line to $f$ at $x$ is $-2 x + 4.$ Find $f (5).$

In problems #9–10, find the function $f$ that satisfies the given conditions.

1. $f''(x) = \sin x - e^{-2x}$ with $f'(0) = \frac{5} {2}$ and $f(0) = 0$
2. $f''(x) = \frac{1} {\sqrt{x}}$ with $f'(4) = 7$ and $f(4) = 25$

1. $f(x) = e^{2x} - \frac{4} {3} x^{\frac{3} {2}} + C$
2. $f(x) = -\cos x + \frac{1} {e^x} + C$
3. $f(x) = \frac{8} {15} x^{\frac{5} {2}} + \frac{4} {15} x^{\frac{7} {2}} + C_1x + C_2$
4. $f(x) = x^6 - \frac{4} {3} x^3 + \frac{7} {3} x + 2$
5. $f(x) = x^3 + \frac{e^{2x}} {2} + \frac{5} {2}$
6. $f(x) = \frac{3} {5} \sqrt[3]{x^5} + \frac{1} {x} + \frac{7} {5}$
7. $f(x) = 2\sin x + \cos x$
8. $f(x) = -x^2 + 4 x + 16$; $f (5) = 11$
9. $f(x) = -\sin x - \frac{1} {4} e^{-2x} + 3x + \frac{1} {4}$
10. $f(x) = \frac{4} {3} x^{\frac{3} {2}} + 3x + \frac{7} {3}$

Feb 23, 2012

Sep 01, 2014