Differential Equation

An equation containing one or more derivatives of a function is a differential equation. Like algebraic equations that describe relations among varying quantities, differential equations describe relations among changing quantities and the rates at which they change. Solving an equation such as equation image indicator $LaTeX: \frac{dy}{dx}=f\left(x\right)$ produces functions $LaTeX: y=F\left(x\right)$ that make the equation true. When solving a differential equation of the form $LaTeX: \frac{dy}{dx}=f\left(x\right)$ , it is convenient to write it using its equivalent differential form dy = f(x)dx. The general antiderivative y = F(x) + C of the function f(x) is called the general solution of the differential equation equation image indicator $LaTeX: \frac{dy}{dx}=f\left(x\right)$ .

Initial-Value Problems

In many applications of integration, you need to find a particular solution of a differential equation that satisfies a condition of the form y(x₀) = y₀ . If you know the value of y = F(x) for one value of x, namely x₀, then it is possible to find a particular solution of an initial-value problem that satisfies an initial condition. For example, given the differential equation equation image indicator $LaTeX: \frac{dy}{dx}=4x^3-1$ that passes through the point (1, -4), it is possible to find the equation of the particular curve satisfying these conditions.

equation image indicator $LaTeX: F\left(x\right)=x^4-x+C$ (General solution)

F(1) = -4 (Initial condition)

equation image indicator $LaTeX: F\left(1\right)=x^4-x+C$

equation image indicator $LaTeX: -4=\left(1\right)^4-1+C$

equation image indicator LaTeX: -4=C

equation image indicator $LaTeX: F\left(x\right)=x^4-x-4$ (Particular solution)

Rectilinear Motion

Antidifferentiation is particularly useful in analyzing rectilinear motion, the motion of an object moving in a straight line. Recall that if an object has position s = f(t), then the velocity function is v(t) = s'(t). This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is a(t) = v'(t), and thus the velocity function is an antiderivative of the acceleration. If the acceleration and the initial values s(0) and v(0) are known, then the position function can be found by antidifferentiating twice.

Consider a particle that moves in a straight line and has acceleration given by a(t) = t - 2 when s(0) = 1 and v(0)= 3. Find the position function s(t).

Since v'(t) = a(t) = t - 2, antidifferentiation gives v(t) = t²/2 - 2t + C.

v(0)= 0²/2 - 2(0) + C and since v(0) = 3, C = 3.

The velocity of the particle is v(t) = t²/2 - 2t + 3.

Since s'(t) = v(t) = t²/2 - 2t + 3, s is the antiderivative of v.

s(t) = t³/6 - t² + 3t + D

s(0) = 0³/6 - 0² + 3(0) + D and since s(0) = 1, D = 1.

The position function is s(t) = t³/6 - t² + 3t + 1.

Only by solving differential equations given initial conditions can many real-life phenomena be usefully described.

Differential Equations, Initial Value, and Rectilinear Motion: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of Differential Equations, Initial Value, and Rectilinear Motion problems.

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