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Section 8.1 Intro to Differential Equations

Find a function \(y\) with the property:

  1. \begin{equation*} y' = y \text{ or } \dfrac{dy}{dx} = y \end{equation*}

    Note 8.1.1.
    \(y = e^x\) works, in fact \(y = C \cdot e^x\) works for any constant \(C\text{.}\) This is called the "general" solution.

  2. \begin{equation*} y'' = -y \text{ or } \dfrac{d^2y}{dx^2} = -y \end{equation*}

    Note 8.1.2.
    \(y = \sin(x)\) works -- so does \(y = \cos(x)\text{.}\) The general solution for this eqation is
    \begin{equation*} y = A\sin(x) + B\cos(x) \end{equation*}
    for any constant \(A\) and \(B\text{.}\)

A differential equation is an equation involving one or more derivatives for an unknown function.

The order of a differential equation is the order of the highest dervative it contains.

Table 8.1.3.
D.E. Order
\(\dfrac{dx}{dy} - 4xy = 3x^5\) 1
\(y'' + 2xy' - e^x = 0\) 2
\(\dfrac{d^3y}{dx^3} + 4x\dfrac{d^2y}{dx^2} + \cos(x)\dfrac{dy}{dx} + y = 7\) 3

A function \(y = y(x)\) is called a solution of a differential equation if the equation is satisfied when \(y\) and its derivatives are substituted into the equation. Example: check that \(y = e^{5x}\) is a solution to the DE \(\dfrac{dy}{dx} - 4y = e^{5x}\text{.}\)

\begin{align*} y \amp = e^{5x}\\ \dfrac{dy}{dx} \amp = 5e^{5x} \end{align*}
\begin{align*} \dfrac{dy}{dx} - 4y \amp = e^{5x}\\ 5e^{5x} - 4(e^{5x}) \amp = e^{5x}\\ e^{5x} \amp = e^{5x} \end{align*}
DE's have lots of real-world applications. Any time we are talking about "rates of change", we are using DE's.

Subsection 8.1.1 Initial Value Problems

Solve the following DE

\begin{equation*} \dfrac{dy}{dx} = 2x - 3 \end{equation*}

subject to the condition that \(y(5) = 17\) (this is known as the initial condition).

  • integrate both sides with respect to x (this is the gereral solution)

    \begin{equation*} y = x^2 - 3x + c \end{equation*}

  • we need \(y(5) = 17\) so we can rewrite as:

    \begin{align*} y(5) \amp = 5^2 - 3(5) + C\\ 17 \amp = 25 - 15 + C\\ 17 \amp = 10 + C\\ C \amp = 7 \end{align*}

  • so, \(y(x) = x^2 - 3x + 7\)

An initial value problem is just a DE with an initial condition.

Subsection 8.1.2 Seperable Differential Equations

A first order DE is said to be seperable if it can be written in the form

\begin{equation*} \dfrac{dy}{dx} = G(x)H(y) \end{equation*}

for some functions \(G\) and \(H\text{.}\)

Seperable DE's are nice because there is a simple method we can use to find the solution.

Example 8.1.1.

Solve the following seperable DE:

\begin{align*} \dfrac{dy}{dx} \amp = - 4 xy^2\\ \dfrac{1}{y^2} \cdot \dfrac{dy}{dx} \amp = - 4x\\ \dfrac{1}{y^2} dy \amp = -4xdx\\ \int \dfrac{1}{y^2} dy \amp = \int -4xdx \\ \int y^{-2} dy \amp = - \int 4x dx \\ \dfrac{y^{-1}}{-1} \amp = \dfrac{-4x^2}{2} + C\\ -\dfrac{1}{y} \amp = -2x^2 + C\\ y = \dfrac{1}{2x^2 - C} \end{align*}

Now solve for the initial condition \(y(0) = 1\text{:}\)

\begin{align*} y(x) \amp = \dfrac{1}{2x^2 - C} \\ y(0) \amp = \dfrac{1}{2(0)^2 - C}\\ y(0) \amp = \dfrac{1}{-C}\\ 1 \amp = -\dfrac{1}{C}\\ C \amp = -1 \end{align*}

So the final answer is \(y(x) = \dfrac{1}{2x^2 + 1}\)