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1.2 Integrals as General and Particular Solutions

first-order differential eq:

\[ \frac{dy}{dx} = f(x, y) \]

Generally contain \( x \) and \( y \)

  • \( x \) is the independent variable
  • \( y \) is the dependent variable

e.g.

\[ \frac{dy}{dx} = xy^2 \quad \text{solution: } y = ? \]

We will start with the case where the right side contains no \( y \):

\[ \frac{dy}{dx} = f(x) \rightarrow \text{just calculus} \]

\[ y = \int f(x) dx \]

e.g.

\[ \frac{dy}{dx} = 2x \]
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\[ y = \int 2x dx = x^2 + C \]

infinitely many solutions

General solution

Graph of parabolas y = x^2 + 3, y = x^2, and y = x^2 - 4 passing through (0,3), (0,0), and (0,-4).

Altogether they are the general solution

Each one is called a particular solution

Finding \( C \) is equivalent to specifying a point the solution must go through. For example, solution to

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Initial Value Problems and Solution Curves

\[ \frac{dy}{dx} = 2x, \quad y(0) = 3 \]

initial or side condition: \( y(0) = 3 \)

Solution is \( y = x^2 + 3 \)

Solution curves do not have common points

Example

\[ \frac{dy}{dx} = x \sqrt{x^2 + 1}, \quad y(0) = 2 \]

initial condition: Given \( \rightarrow \) "initial value problem" or IVP

\[ y = \int x \sqrt{x^2 + 1} \, dx \]

Let \( u = x^2 + 1 \)

\( du = 2x \, dx \)

\[ = \int \frac{1}{2} u^{1/2} \, du = \frac{1}{2} \frac{u^{3/2}}{3/2} + C = \frac{1}{3} u^{3/2} + C \]
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\[ y = \frac{1}{3} (x^2 + 1)^{3/2} + C \quad \text{(general solution)} \]

Use the given initial condition \( y(0) = 2 \) to find \( C \)

\[ 2 = \frac{1}{3} (0 + 1)^{3/2} + C \quad \text{so} \quad C = \frac{5}{3} \]

\[ y = \frac{1}{3} (x^2 + 1)^{3/2} + \frac{5}{3} \]

Second-order Equations

Second-order: \[ \frac{d^2y}{dx^2} = f(x) \]

\[ \frac{dy}{dx} = \int f(x) \, dx + C_1 = F(x) + C_1 \]

\[ y = \int F(x) \, dx + C_1x + C_2 \]

Two constants: \( C_1 \) and \( C_2 \)

Two initial/side conditions to find constants

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\(n^{\text{th}}\)-order \(\rightarrow n\) initial conditions

\[y(x_0) = y_0, \quad y'(x_0) = y'_0, \quad \text{etc.}\]

Example

\[y'' = -10\]\[y' = -10x + C_1\]\[y = -5x^2 + C_1x + C_2\]

need two conditions: for example \(y(0) = 0\) and \(y'(0) = 1\)

\[1 = -10(0) + C_1 \rightarrow C_1 = 1\]\[y = -5x^2 + x + C_2\]\[y(0) = 0 \rightarrow 0 = -5(0)^2 + (0) + C_2 \quad \text{so } C_2 = 0\]
\[y = -5x^2 + x\]
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\(y'' = -10\) can be interpreted physically

let's use \(t\) instead of \(x\) as the independent variable

if \(y\) is the position of an object

then \(y'\) is velocity and \(y''\) is the acceleration

and \(y(0) = 0 \rightarrow\) initial position: at \(t = 0\), \(y\) is at 0

\(y'(0) = 1 \rightarrow\) initial velocity: at \(t = 0\), \(y'\) is 1

A graph of a downward parabola starting at the origin (0,0) on y vs t axes. An arrow at the origin indicates initial slope.
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\[ y'' + 2y' + y = 0 \]

can be interpreted physically as a mass-spring-damper problem

A mass-spring-damper system diagram with a mass m connected to a wall by a spring and a damper.
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Calculus review

trig subs.

\[ \frac{dy}{dx} = \frac{1}{x^2 + 4} \]

triangle with sides \( x^2+4 \), \( x \), \( 2 \)

Right triangle with angle theta, opposite side x, adjacent side 2, and hypotenuse x squared plus 4.

relate \( x \) and \( \theta \):

\[ \tan \theta = \frac{x}{2} \]\[ x = 2 \tan \theta \]\[ dx = 2 \sec^2 \theta \, d\theta \]
\[ y = \int \frac{1}{x^2 + 4} \, dx \]\[ = \int \frac{1}{4 \tan^2 \theta + 4} \cdot 2 \sec^2 \theta \, d\theta = \int \frac{1}{4 \sec^2 \theta} \cdot 2 \sec^2 \theta \, d\theta = \int \frac{1}{2} \, d\theta \]

Note: \( 4(\tan^2 \theta + 1) = 4 \sec^2 \theta \)

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Substitution and Final Solution

\[ y = \frac{1}{2} \theta + c \]

Given the trigonometric relationship:

\[ \tan \theta = \frac{x}{2} \implies \theta = \tan^{-1} \left( \frac{x}{2} \right) \]

Substituting \( \theta \) back into the equation for \( y \), we obtain the final result:

\[ y = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + c \]