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Section 12.4 Initial Value Problems (IN4)

Subsection 12.4.1 Activities

Note 12.28.

In this section we will discuss the relationship between antiderivatives and solving simple differential equations. A differential equation is an equation that has a derivative. For this section we will focus on differential equations of the form
\begin{equation*} \frac{dy}{dx} = f(x). \end{equation*}
Our goal is to find a relationship of \(y(x)\) that satisfies the differential equation. We can solve for \(y(x)\) by finding the antiderivative of \(f(x)\text{.}\)

Activity 12.29.

Which of the following equations for \(y(x)\) satisfies the differential equation
\begin{equation*} \frac{dy}{dx} = x^2+2x\text{.} \end{equation*}
  1. \(\displaystyle y(x) = \frac{x^3}{3} + x^2 + 4\)
  2. \(\displaystyle y(x) = 2x + 2\)
  3. \(\displaystyle y(x) = \frac{x^3}{3} + x^2 + 10\)
  4. \(\displaystyle y(x) = \frac{x^3}{3} + x^2\)
  5. \(\displaystyle y(x) = 2x \)

Remark 12.30.

In Activity 12.29 there are more than one solution that satisfies the differential equation. In fact their is a family of functions that satisfies the differential equation, that is
\begin{equation*} f(x) = \frac{x^3}{3} + x^2 + c_1, \end{equation*}
where \(c_1\) is an arbitrary constant yet to be defined. To find \(c_1\) we have to have some initial value for the differential equation, \(y(x_0) = y_0\text{,}\) where the point \((x_0,y_0)\) is the starting point for the differential equation. In general this section we will focus on solving initial value problems (differential equation with an initial condition) of the form,
\begin{equation*} \frac{dy}{dx} = f(x), \;\;\; y(x_0) = y_0. \end{equation*}

Activity 12.31.

Which of the following equations for \(y(x)\) satisfies the differential equation and initial condition,
\begin{equation*} \frac{dy}{dx} = x^2+2x, \,\,\,y(3) = 16. \end{equation*}
  1. \(\displaystyle y(x) = \frac{x^3}{3} + x^2 - 4\)
  2. \(\displaystyle y(x) = \frac{x^3}{3} + x^2 + 2\)
  3. \(\displaystyle y(x) = \frac{x^3}{3} + x^2 -2\)
  4. \(\displaystyle y(x) = \frac{x^3}{3} + x^2 + 16\)

Activity 12.32.

Which of the following functions satisfies the initial value problem,
\begin{equation*} \frac{dy}{dx} = \sin(x), \,\,\, y(0) = 1. \end{equation*}
  1. \(\displaystyle y(x) = \cos(x)\)
  2. \(\displaystyle y(x) = \cos(x) + 2\)
  3. \(\displaystyle y(x) = \cos(x) + 1\)
  4. \(\displaystyle y(x) = -\cos(x)\)
  5. \(\displaystyle y(x) = -\cos(x) + 2 \)

Activity 12.33.

One of the applications of initial value problems is calculating the distance traveled from a point based on the velocity of the object. Given that the velocity of the of an object in km/hr is approximated by \(v(t) = \cos(t) + 1\text{,}\) what is the approximate distance traveled by the object after 1 hour?
  1. \(s(1) \approx 1\) km
  2. \(s(1) \approx 0.1585\) km
  3. \(s(1) \approx 1.8415\) km
  4. \(s(1) \approx 2.3415\) km

Activity 12.34.

So far we have only been going from velocity to position of an object. Recall that to find the acceleration of an object, you can take the derivative of the velocity of an object. Let use say we have the acceleration of a falling object in m/s\(^2\) given by \(a(t) = -9.8\text{.}\)
(a)
What is the velocity of the falling object, if the initial velocity is given by \(v(0) = 0\) m/s.
  1. \(v(t) = -9.8t\) m
  2. \(v(t) = -9.8t\) m/s
  3. \(v(t) = 9.8t\) m/s
  4. \(v(t) = 9.8t+1\) m
(b)
What is the position of the object, if the initial position is given by \(s(0) = 10\) m.
  1. \(s(t) = 4.9t + 10\) m
  2. \(s(t) = -4.9t^2 + 14.9\) m
  3. \(s(t) = -4.9t^2 + 10\) m
  4. \(s(t) = 4.9t + 5.1\) m

Activity 12.35.

Let \(f'(x)=-12 \, x - 6 \text{.}\) Find \(f(x)\) such that \(f(5)=-179\text{.}\)

Subsection 12.4.2 Videos

Figure 143. Video for IN4

Subsection 12.4.3 Exercises

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