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PAGE 1

2.2 (continued)

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

if \( f(x, y) \) can be expressed as a function of \( \frac{y}{x} \), then the equation is called homogeneous → has many meanings in diff. eqs.

a homogeneous eq. can be turned into a separable eq. by a change of variable

Example

for example, \( \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy} \) NOT separable as is

we can rewrite it by dividing by \( x^2 \) on top & bottom

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

function of \( \frac{y}{x} \) on the right

homogeneous

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define new variable \( v = \frac{y}{x} \) so \( y = vx \)

rewrite \( \frac{dy}{dx} = \frac{1 + 3(\frac{y}{x})^2}{2(\frac{y}{x})} \) in terms of \( v \) and \( x \)

\[ y = vx \quad \text{so} \quad \frac{dy}{dx} = v + x \frac{dv}{dx} \]

function of \( x \)

product rule

Equation Transformation

eq. becomes

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

Separable in \( v \) and \( x \)

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

integrate

\[ \ln |1+v^2| = \ln |x| + C \]
\[ 1+v^2 = e^C \cdot e^{\ln |x|} \]
\[ = e^C \cdot |x| = \underbrace{\pm e^C}_{C} \cdot x \]
\[ = Cx \]
\[ v^2 = Cx - 1 \]
\[ \frac{y^2}{x^2} = Cx - 1 \]
\[ y^2 = Cx^3 - x^2 \]
general solution
(implicit)
PAGE 4

if \( \frac{dy}{dx} = f(x,y) \) is homogeneous,

then \( \frac{dy}{dx} = g(\frac{y}{x}) \)

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

so slope only depends on \( \frac{y}{x} \)

so any line through origin (constant \( \frac{y}{x} \)) has the same slope everywhere

\[ \frac{dy}{dx} = \frac{1 + 3(\frac{y}{x})^2}{2(\frac{y}{x})} \]
A Cartesian coordinate system with lines through the origin, marked with a large red 'WRONG' across it.
PAGE 5

Redo Correctly

\[ \frac{dy}{dx} = \frac{1 + 3(\frac{y}{x})^2}{2(\frac{y}{x})} \]
A slope field on a Cartesian coordinate system showing symmetry with respect to the origin.

\( (\frac{y}{x} = 1) \)

\[ \frac{dy}{dx} = \frac{1+3}{2} = 2 \]

\( (\frac{y}{x} = -1) \)

\[ \frac{dy}{dx} = \frac{1+3}{-2} = -2 \]

Slope field is symmetric with respect to the origin

PAGE 6

Solving Homogeneous Equations

ALL homogeneous eqs. can be solved that way

NOT always necessary

for example,

\[ \frac{dy}{dx} = \frac{y}{x} \] is separable
is homogeneous
\[ y' = \frac{1}{x} y \]
\[ y' - \frac{1}{x} y = 0 \] is linear
PAGE 7

2.3 Modeling with First-Order Diff. Eqs.

mixing problem example

A tank initially contains 40 kg of salt dissolved in 600 L of water.

Water containing \( \frac{1}{2} \) kg of salt per liter is poured into the tank at the rate of 4 L/min.

The well-stirred solution is let out at the same rate (4 L/min).

How much salt is in the tank at any given time?

A cylindrical tank with an arrow pointing in labeled concentration 1/2 kg/L and flow rate 4 L/min.
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define \( Q(t) \) as amount of salt in the tank at time \( t \) (kg)

\[ \frac{dQ}{dt} = (\text{rate of salt in}) - (\text{rate of salt out}) \]
\[ = \underbrace{(\frac{1}{2} \text{ kg/L})}_{\text{concentration}} \underbrace{(4 \text{ L/min})}_{\text{flow rate}} - \underbrace{(? \text{ kg/L})}_{\text{concentration}} \underbrace{(4 \text{ L/min})}_{\text{flow rate}} \]

concentration out: concentration of salt water in tank

\[ \frac{\text{salt in kg}}{\text{volume in L}} = \frac{Q}{600} \]

(4 in, 4 out)

\[ \frac{dQ}{dt} = 2 - \frac{Q}{600} \cdot 4 = 2 - \frac{1}{150} Q \]
  • linear
  • separable

let's solve as separable

PAGE 9

Solving a First-Order Differential Equation

We begin with the differential equation for the quantity \( Q \) with respect to time \( t \):

\[ \frac{dQ}{dt} = \frac{1}{150} (300 - Q) \]

Separating the variables, we get:

\[ \frac{1}{300 - Q} dQ = \frac{1}{150} dt \]

Integrating both sides:

\[ -\ln |300 - Q| = \frac{1}{150} t + C \]

Multiplying by -1 and adjusting the constant:

\[ \ln |300 - Q| = -\frac{1}{150} t + C \]

Exponentiating both sides to solve for \( Q \):

\[ 300 - Q = C e^{-\frac{1}{150} t} \]

Rearranging for \( Q(t) \):

\[ Q(t) = 300 - C e^{-\frac{1}{150} t} \]

Applying Initial Conditions

Given the initial condition \( Q(0) = 40 \):

\[ 40 = 300 - C \]

Solving for \( C \):

\[ C = 260 \]

So, the final particular solution is:

\[ Q(t) = 300 - 260 e^{-\frac{1}{150} t} \]