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Solving a Bernoulli Differential Equation

Supp. C — Solving the differential equation for \(x > 0\):

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

Let us use the substitution:

\[u = y^3\]

Differentiating both sides with respect to \(x\):

\[\frac{du}{dx} = 3y^2 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{3y^2} \frac{du}{dx}\]

Substituting these into the original equation:

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

Linear Form

Multiply by 3 to put the equation in standard linear form:

\[u' + \frac{3}{x} u = \frac{6}{x^2}\]

Integrating Factor

Calculate the integrating factor \(\mu\):

\[\mu = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = e^{\ln x^3} = x^3\]

Multiplying the linear equation by \(\mu = x^3\):

\[x^3 u' + 3x^2 u = 6x\]
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Continuing from the previous page, we recognize the left side as a derivative:

\[\frac{d}{dx} (x^3 u) = 6x\]

Integrating both sides with respect to \(x\):

\[x^3 u = 3x^2 + C\]

Solving for \(u\):

\[u = \frac{3}{x} + \frac{C}{x^3}\]

Substituting back \(u = y^3\) to find the final solution:

\[y^3 = \frac{3}{x} + \frac{C}{x^3}\]

Problem 32: Homogeneous Equation

Given the differential equation:

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

Using the substitution for homogeneous equations:

\[y = xv \implies v = \frac{y}{x}\]

Divide numerator and denominator by \(x^2\):

\[\frac{dy}{dx} = \frac{\frac{x^2 + 3y^2}{x^2}}{\frac{2xy}{x^2}} = \frac{1 + 3v^2}{2v}\]
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Solving Homogeneous Differential Equations

Using the substitution \( y = xv \), we find the derivative:

\[ \frac{dy}{dx} = \frac{d}{dx}(xv) = x \frac{dv}{dx} + v \]

New Differential Equation

Substituting into the original DE:

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

Isolating the derivative term:

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

Separation of Variables

Rearranging to separate \( v \) and \( x \):

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

Integrating both sides:

\[ \ln|1 + v^2| = \ln|x| + K \]

Exponentiating both sides:

\[ 1 + v^2 = e^{\ln|x| + K} = e^{\ln x} \cdot e^K \]
\[ 1 + v^2 = Cx \]
\[ v^2 = Cx - 1 \]

Final Solution

Substituting back \( v = \frac{y}{x} \):

\[ \frac{y^2}{x^2} = Cx - 1 \]
\[ y^2 = Cx^3 - x^2 \]
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Problem 22: Initial Value Problem and Vertical Tangents

Given the differential equation:

\[ y' = \frac{3x^2}{3y^2 - 4} \]

with initial condition: \( y(1) = 0 \)

Implicit Solution

The solution to the differential equation is found to be:

\[ y^3 - 4y = x^3 - 1 \]

Question: Over what interval is this solution valid?

Vertical Tangents

The derivative \( y' \) is undefined when the denominator is zero:

\[ 3y^2 - 4 = 0 \implies y^2 = \frac{4}{3} \implies y = \pm 1.1547 \]

To find the corresponding \( x \) values, substitute \( y \) back into the implicit solution:

\[ x^3 = 1 + (y^3) - 4(y) \]

For \( y = -1.1547 \):

\[ x^3 = 1 + (-1.1547)^3 - 4(-1.1547) \]\[ x \approx -1.28 \]
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2.3 Modeling with First Order Equations

Example 1. A 200-liter water tank initially contains 100 L of pure water. A mixture containing a concentration of 2 g/L of salt enters the tank at a rate of 3 L/min, and the well-stirred mixture leaves the tank at the same rate. Find an expression for the amount of salt in the tank at any time.

let \( y(t) \) be amount of salt (in grams)

need a DE

\[ \frac{dy}{dt} = (\text{rate in}) - (\text{rate out}) \]\[ \frac{dy}{dt} = \underbrace{(2 \text{ g/L})(3 \text{ L/min})}_{\text{in}} - \underbrace{\left( \frac{y \text{ g}}{100 \text{ L}} \right)(3 \text{ L/min})}_{\text{out}} \]
\[ \frac{dy}{dt} = 6 - 0.03y \quad \text{linear and separable} \]\[ \frac{1}{6 - 0.03y} dy = dt \]\[ \vdots \]\[ y = 200 + Ce^{-0.03t} \]
\[ y = 200 - 200e^{-0.03t} \]

IC: \( y(0) = 0 \)

"pure" water

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\[ \lim_{t \to \infty} y = 200 \]

tank has same concentration as flow in as \( t \to \infty \)

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Example 2: Mixing Tank with Variable Volume

Same set-up as in example 1. But what if the well-stirred mixture is let out of the tank at a rate of 2 L/min instead of 3 L/min? What is the amount of salt in the tank right before the moment the tank overflows?

\[ y(t) : \text{ salt in tank (in grams)} \]

\[ \frac{dy}{dt} = \underbrace{(2 \text{ g/L})(3 \text{ L/min})}_{\text{in}} - \underbrace{\left( \frac{y}{100+t} \right)(2)}_{\text{out}} \]

Note on Volume:

Net gain of 1 L/min

  • 5 min \(\rightarrow\) 105 L
  • 15 min \(\rightarrow\) 115 L
  • t min \(\rightarrow\) (100 + t) L

\(100+t\) is the volume of water.

Solving the Differential Equation

The equation is a first order linear differential equation:

\[ y' = 6 - \frac{2}{100+t}y \]\[ y' + \frac{2}{100+t}y = 6 \]

Integrating Factor

\[ \mu = e^{\int \frac{2}{100+t} dt} \]\[ \mu = e^{2 \ln(100+t)} \]\[ \mu = (100+t)^2 \]

Initial Condition and Solution

IC: \( y(0) = 0 \)

\[ y(t) = 2(100+t) - 200(100)^2(100+t)^{-2} \]

Overflow Analysis

Overflows when \( t > 100 \)

\[ y(100) = 350 \text{ g} \]\[ \text{Concentration} = \frac{350 \text{ g}}{200 \text{ L}} = 1.75 \text{ g/L} \]
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Example 3: Projectile Motion with Air Resistance

An object of mass 1 kg is tossed straight up from ground level with an initial velocity of 5 m/s. Force due to air resistance can be modeled as \( \frac{|v|}{20} \), in the direction opposite to the velocity. Find expressions for the velocity and height as a functions of time.

define positive direction

y positive up

drag changes direction!

ground \( y = 0 \)

Diagram of an object's vertical path from a ground line labeled y=0.

upward part:

Free body diagram for upward motion with gravity and drag both pointing down.

up part: \( F = ma \)

\[ m \frac{dv}{dt} = -mg - \frac{|v|}{20} \]\[ \frac{dv}{dt} = -g - \frac{v}{20m} \]

\( v > 0 \) going up so \( |v| = v \)

downward part:

Free body diagram for downward motion with gravity down and drag up.
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down part: \( F = ma \)

\[ m \frac{dv}{dt} = -mg + \frac{|v|}{20} \]\[ \frac{dv}{dt} = -g - \frac{v}{20m} \]

\( v < 0 \) going down

\( |v| = -v \)

e.g. \( v = -2 \)

\( |-2| = -(-2) = 2 \)

so one DE for both parts

\[ \frac{dv}{dt} = -g - \frac{v}{20m} \]

linear and separable

\( m = 1 \)

\( g = 9.8 \)

\( v(0) = 5 \)

\[ v(t) = -20mg + (5 + 20mg)e^{-\frac{1}{20m}t} \]

\[ v(t) = -196 + 201e^{-0.05t} \] velocity

\[ y(t) = \int v(t) dt \]\[ y(0) = 0 \text{ (ground)} \]

\[ y(t) = -196t - 4020e^{-0.05t} + 4020 \]

when does max height occur?

\( v = 0 \)

max height: \( y(t^*) \)

solve for \( t \): \( t \approx 0.504 \)

when does it hit ground?

solve \( y = 0 \rightarrow 2 \text{ values of } t \)

\( \hookrightarrow \text{2nd } t \)