PAGE 1

1st-Order Homogeneous Diff. Eqs.

A function \( f(x, y) \) is said to be homogeneous of degree \( n \) if

\[ f(\lambda x, \lambda y) = \lambda^n f(x, y) \]

Concept Note: Scaling all inputs by an arbitrary non-zero factor \( \lambda \) factors out the constant raised to the power \( n \), which denotes the homogeneity degree of the function.

For example,

\[ f(x, y) = x + y \] \[ f(\lambda x, \lambda y) = \lambda x + \lambda y = \lambda(x + y) \quad \text{homogeneous of degree } 1 \]

\[ g(x, y) = xy \] \[ g(\lambda x, \lambda y) = (\lambda x)(\lambda y) = \lambda^2 xy \quad \text{homogeneous deg } 2 \]

A 1st-order diff. eq. is said to be homogeneous if it is of the form

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

such that \( M \) and \( N \) are homogeneous of the same degree

Solving Strategy: When \( M(x, y) \) and \( N(x, y) \) possess the same degree \( n \), the ratio \( \frac{M(x, y)}{N(x, y)} \) can be written purely as a function of \( \frac{y}{x} \). This enables the standard substitution \( y = v x \) (or \( v = \frac{y}{x} \)), which transforms the equation into a separable first-order differential equation in terms of \( v \) and \( x \).

PAGE 2

Form of Homogeneous Differential Equations

in such a case, the eq. can always be written as

\[ \frac{dy}{dx} = \frac{M(x,y)}{N(x,y)} = f\left(\frac{y}{x}\right) \]

Key Principle: When both \(M(x,y)\) and \(N(x,y)\) are homogeneous functions of the exact same degree, dividing the numerator and denominator by \(x^n\) reduces the entire right-hand side to an expression purely in terms of the ratio \(\frac{y}{x}\).

First Example

for example,

\[ \frac{dy}{dx} = \frac{x+y}{x} \]\[ = \frac{x}{x} + \frac{y}{x} \]\[ \frac{dy}{dx} = 1 + \frac{y}{x} = f\left(\frac{y}{x}\right) \]

numerator: homogeneous deg \(1\)

denominator: homogeneous deg \(1\)

Analysis: Here, both terms in the numerator (\(x\) and \(y\)) and the denominator (\(x\)) have degree \(1\). Dividing term-by-term immediately reveals the function \(f(u) = 1 + u\), where \(u = \frac{y}{x}\).

Second Example

another example:

\[ \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy} \]\[ = \frac{\frac{x^2 + 3y^2}{x^2}}{\frac{2xy}{x^2}} \]

numerator: homogeneous deg \(2\)

denominator: homogeneous deg \(2\)

Next Steps: Dividing through by \(x^2\) transforms the expression into \(\frac{1 + 3\left(\frac{y}{x}\right)^2}{2\left(\frac{y}{x}\right)}\), which is of the form \(f\left(\frac{y}{x}\right)\), setting it up for the standard substitution \(v = \frac{y}{x}\).

PAGE 3

Homogeneous Differential Equations: Substitution Method

Page 3

\[ = \frac{1 + 3\left(\frac{y}{x}\right)^2}{2\left(\frac{y}{x}\right)} = f\left(\frac{y}{x}\right) \]

\[ \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy} \]

is not linear, not separable.

How to solve?

It is homogeneous, so the right side is \( f\left(\frac{y}{x}\right) \).

\( \rightarrow \) Suggesting a change of variable may help.

\[ \frac{dy}{dx} = \frac{1 + 3\left(\frac{y}{x}\right)^2}{2\left(\frac{y}{x}\right)} \]

Let \( v = \frac{y}{x} \) \(\quad\text{so}\quad\) \( y = vx \)

Concept Note:

A first-order ordinary differential equation \( \frac{dy}{dx} = f(x, y) \) is homogeneous of degree zero if \( f(x, y) \) can be written purely as a function of the single ratio \( \frac{y}{x} \). Substituting \( v = \frac{y}{x} \) transforms it into a separable equation for \( v(x) \).

\[ \frac{d\mathbf{y}}{dx} = \frac{1 + 3v^2}{2v} \]

We want eq. in terms of \( v \), the new variable.

Eliminate \( y \) from \( y = vx \).

Next Step Breakdown:

Since \( y = vx \), apply the product rule with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx}(vx) = v + x\frac{dv}{dx} \). Substituting this into the left-hand side allows complete elimination of \( y \).

PAGE 4

Substitution and Reduction to Separable Form

\[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}\left(\overbrace{v}^{\text{function of } x} x\right) \\[1ex] &= v + x \frac{dv}{dx} \quad \text{left side of new eq.} \end{aligned} \]

Product Rule Application: Since \( y = vx \) and \( v \) is treated as an unknown function of \( x \), differentiating both sides with respect to \( x \) yields \( \frac{d}{dx}(vx) = \frac{dv}{dx}\cdot x + v\cdot 1 = v + x\frac{dv}{dx} \).

\[ v + x \frac{dv}{dx} = \frac{1 + 3v^2}{2v} \]

\[ \begin{aligned} x \frac{dv}{dx} &= \frac{1 + 3v^2}{2v} - v \\[1ex] &= \frac{1 + 3v^2}{2v} - \frac{2v^2}{2v} \end{aligned} \]

\[ x \frac{dv}{dx} = \frac{1 + v^2}{2v} \]

new eq. in \( v \) and \( x \)

note it is separable in \( v \) and \( x \)

PAGE 5

Separable Differential Equation: Integration & Back-Substitution

1. Separation of Variables & Integration

Starting from the separated differential form:

\( \frac{2v\,dv}{1 + v^2} = \frac{1}{x}\,dx \)

Integrate both sides:

\( \int \frac{2v}{1 + v^2}\,dv = \int \frac{1}{x}\,dx \)

Integration Technique: On the left side, notice that the numerator \( 2v \) is the exact derivative of the denominator \( 1 + v^2 \). Using substitution \( u = 1 + v^2 \), \( du = 2v\,dv \), the integral evaluates directly to \( \ln|1 + v^2| \).

\( \ln |1 + v^2| = \ln |x| + C \)

2. Exponentiation & Simplifying the Constant

Exponentiate both sides to eliminate the natural logarithm:

\[ \begin{aligned} 1 + v^2 &= e^{\ln |x| + C} = e^C e^{\ln |x|} \\[6pt] &= e^C \cdot |x| \\[6pt] 1 + v^2 &= \underbrace{\pm e^C}_{C} \cdot x \end{aligned} \]

Consolidating Constants: Because \( |x| = \pm x \), we combine \( \pm e^C \) into a single arbitrary constant \( C \). This simplifies the relation to \( 1 + v^2 = Cx \).

\[ 1 + v^2 = C x \]

Undo substitution: \( v = \frac{y}{x} \)

3. Back-Substitution & Final Solution

Substitute \( v = \frac{y}{x} \) back into the equation:

\[ \frac{y^2}{x^2} = C x - 1 \]

Multiply both sides by \( x^2 \) to isolate \( y^2 \):

\[ y^2 = C x^3 - x^2 \]

PAGE 6

Homogeneous Equations

Homogeneous eq: \[ \frac{dy}{dx} = f\left(\frac{y}{x}\right) \]

Slope depends on \( \frac{y}{x} \)

which means along any line \( y = mx \), the slope is constant

Conceptual Note: In a homogeneous first-order differential equation, the derivative \( \frac{dy}{dx} \) depends solely on the ratio \( \frac{y}{x} \). Since \( \frac{y}{x} = m \) is constant along any radial ray from the origin \( y = mx \), the direction field has identical slopes along those lines.

Summary of homogeneous eq.

Rewrite \( \frac{dy}{dx} = \frac{M(x,y)}{N(x,y)} \) as \( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)
Define \( v = \frac{y}{x} \rightarrow y = vx \rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Rewrite eq. in terms of \( v \) and \( x \) \( \Longrightarrow \) separable / linear / both
Solve for \( v \), undo \( v = \frac{y}{x} \) for \( y \)

Derivation of the Derivative: By using the product rule on \( y(x) = v(x) \cdot x \), we obtain \( \frac{dy}{dx} = \frac{dv}{dx} \cdot x + v \cdot 1 = v + x\frac{dv}{dx} \). Substituting this into \( \frac{dy}{dx} = f(v) \) yields \( v + x\frac{dv}{dx} = f(v) \), which separates directly into \( \frac{dv}{f(v) - v} = \frac{dx}{x} \).

\[ \frac{dy}{dx} = \frac{y}{x} \]

\[ \frac{dy}{dx} - \frac{1}{x}y = 0 \]

choose method carefully!

separable, linear, homogeneous

PAGE 7

Slope Field of \(\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}\)

PAGE 8

Application

The Salty Tank

Figure: Diagram of a cylindrical mixing tank with inflow and outflow labeled as salt water. Show Details

Schematic diagram of a vertical cylindrical mixing tank modeling the salt amount \( S(t) \). An arrow at the top labeled "Salt water" indicates an inflow rate of \( 3\text{ gal/min} \) carrying \( 2\text{ lbs/gal} \) of salt. An exit arrow at the bottom right labeled "Salt water" illustrates the well-mixed solution draining at \( 3\text{ gal/min} \).

A large tank initially holds \( 100 \) gallons of pure water.

Inflow: A brine solution containing \( 2\text{ lbs} \) of salt per gallon is pumped into the tank at a rate of \( 3\text{ gallons per minute} \).
Mixing: The liquid in the tank is kept well-mixed.
Outflow: The mixture is pumped out of the tank at the same rate of \( 3\text{ gallons per minute} \).

Set up a differential equation for the amount of salt \( S(t) \) (in pounds) in the tank at time \( t \) (in minutes).
Solve the initial value problem to find an explicit formula for \( S(t) \).
Determine the limiting amount of salt in the tank, and explain why this value makes physical sense.

Conceptual Guide: Mixing Problems

The rate of change of the substance in the tank is modeled by the mass balance equation:

\[ \frac{dS}{dt} = \text{Rate}_{\text{in}} - \text{Rate}_{\text{out}} \]

Rate in: \( (\text{concentration in}) \times (\text{flow rate in}) = (2\text{ lb/gal}) \times (3\text{ gal/min}) = 6\text{ lb/min} \).
Volume: Since inflow rate equals outflow rate (\( 3\text{ gal/min} = 3\text{ gal/min} \)), the volume is constant at \( V(t) = 100\text{ gal} \).
Rate out: \( (\text{concentration out}) \times (\text{flow rate out}) = \left(\frac{S(t)}{100}\text{ lb/gal}\right) \times (3\text{ gal/min}) = \frac{3}{100}S(t)\text{ lb/min} \).
Differential Equation: \( \frac{dS}{dt} = 6 - \frac{3}{100}S \), with initial condition \( S(0) = 0 \).
Steady-State Limit: As \( t \to \infty \), \( \frac{dS}{dt} \to 0 \implies S(\infty) = \frac{6}{3/100} = 200\text{ lbs} \), which equals \( 100\text{ gal} \times 2\text{ lb/gal} \).

PAGE 9

Mixing Problem: Rate of Salt in a Tank

\( S(t) \): amount of salt (\(\text{lb}\)) in the tank

\[ \begin{aligned} \frac{dS}{dt} &= (\text{rate of salt in}) - (\text{rate of salt out}) \\[8pt] &= (\text{flow rate in})(\text{concentration in}) - (\text{flow rate out})(\text{conc. out}) \\[8pt] &= \left(3\text{ gal/min}\right)\left(2\text{ lb/gal}\right) - \left(3\text{ gal/min}\right)\left(\frac{\text{amount salt in tank}}{\text{volume of tank}}\right) \\[8pt] &= \left(6\text{ lb/min}\right) - \left(3\text{ gal/min}\right)\left(\frac{S}{100}\text{ lb/gal}\right) \end{aligned} \]

Concept Breakdown: The net rate of change of salt \(\frac{dS}{dt}\) is modeled by inflow rate minus outflow rate. Rate in = \((3\text{ gal/min}) \times (2\text{ lb/gal}) = 6\text{ lb/min}\). Rate out = \((3\text{ gal/min}) \times \left(\frac{S}{100}\text{ lb/gal}\right) = \frac{3}{100}S\text{ lb/min}\), assuming complete mixing and a constant tank volume of \(100\text{ gal}\).

PAGE 10

Salt Amount in Tank Over Time

Figure: coordinate graph showing exponential curve S(t) = 200(1 - e^{-0.03t}) approaching an equilibrium of 200\text{ lbs} — Figure: Figure: coordinate graph showing exponential curve \( S(t) = 200(1 - e^{-0.03t}) \) approaching an equilibrium of \( 200\text{ lbs} \) Show Details

Coordinate plot displaying the salt amount in a tank over time. The horizontal axis measures time \( t \) in minutes from \( 0 \) to \( 200 \) in increments of \( 25\text{ min} \). The vertical axis measures salt \( S(t) \) in pounds from \( 0 \) to \( 200 \) in increments of \( 25\text{ lbs} \). A solid blue curve represents the function \( S(t) = 200(1 - e^{-0.03t}) \), beginning at the origin \( (0, 0) \) and rising monotonically toward a horizontal asymptote. A dashed red horizontal line marks the equilibrium level at \( S = 200\text{ lbs} \). A data point is marked on the curve at \( t = 100\text{ min} \) with an arrow pointing to the annotation \( t = 100\text{ min} \), \( S \approx 190.0\text{ lbs} \).

Governing Equation & Asymptote:

\[ S(t) = 200\left(1 - e^{-0.03t}\right) \]

Equilibrium Level: \( \lim_{t \to \infty} S(t) = 200\text{ lbs} \)

Key values of salt amount S(t) evaluated at specific times t
Time \( t \) (minutes)	Salt \( S(t) \) (pounds)	Status / Milestone
\( 0 \)	\( 0\text{ lbs} \)	Initial amount: \( S(0) = 200(1 - e^{0}) = 0 \)
\( 100 \)	\( \approx 190.0\text{ lbs} \)	Labeled point: \( S(100) = 200(1 - e^{-3}) \approx 190.04\text{ lbs} \)
\( t \to \infty \)	\( 200\text{ lbs} \)	Horizontal asymptote / Equilibrium concentration

Conceptual Note: First-Order Linear Mixing Model

This curve describes the solution to a standard mixing problem differential equation:

\[ \frac{dS}{dt} + 0.03S = 6, \quad S(0) = 0 \]

The exponential term \( e^{-0.03t} \) decays toward zero as time increases, causing the total salt \( S(t) \) in the tank to level off smoothly at the horizontal asymptote of \( 200\text{ lbs} \).

Definition of Homogeneous Function