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1.4 Separable Differential Eqs

revisit
\[ \frac{dy}{dx} = 2x \]

solution is easy: \[ y = \int 2x \, dx = x^2 + c \]

because we can isolate \( x \) on its own side

a separable eq. is one that we can do something to isolate \( x \) from \( y \). Then we integrate both sides.

for example:

\[ \frac{dy}{dx} = xy \]

think of \( \frac{dy}{dx} \) as a fraction (technically not true but the end result is the same)

\[ dy = xy \, dx \]\[ \frac{1}{y} \, dy = x \, dx \]

now integrate both sides

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\[ \int \frac{1}{y} \, dy = \int x \, dx \]\[ \ln |y| + c_1 = \frac{1}{2}x^2 + c_2 \]
\[ \ln |y| = \frac{1}{2}x^2 + (c_2 - c_1) \]

still a constant call it \( C \)

\[ \ln |y| = \frac{1}{2}x^2 + C \]\[ e^{\ln |y|} = e^{\frac{1}{2}x^2 + C} \]\[ |y| = e^{\frac{1}{2}x^2} \cdot e^C \]

still a constant call it \( C \) (we abuse the letter \( C \))

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

explicit form

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Example: Solving a Separable Differential Equation

\[ (1+x)y' = y \]
\[ (1+x) \frac{dy}{dx} = y \]

Use multiplication or division to separate \(x\) and \(y\).

\[ \frac{1}{y} dy = \frac{1}{x+1} dx \]
\[ \int \frac{1}{y} dy = \int \frac{1}{1+x} dx \]
\[ \ln |y| = \ln |1+x| + C \]
\[ e^{\ln |y|} = e^{\ln |1+x|} \cdot e^C \]
\[ |y| = C |1+x| \]
\[ y = C(1+x) \]

\(C\) depends on initial condition

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Non-Separable Differential Equations

Not all differential eqs. are separable.

For example,

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

Can't separate \(x\) and \(y\) by multiplication or division.

Never do addition/subtraction

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

?

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Example: Solving a Differential Equation

Given the initial value problem:

\[ \frac{dy}{dx} = xy^2 - y^2, \quad y(1) = 2 \]

First, factor the right side to separate variables:

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

Separate the variables and integrate both sides:

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

Perform the integration:

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

General Solution

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

Finding the Particular Solution

Apply the initial condition \( y(1) = 2 \), which means when \( x = 1 \), \( y = 2 \):

\[ 2 = \frac{-1}{\frac{1}{2}(1)^2 - 1 + C} = \frac{-1}{-\frac{1}{2} + C} \]

Solving for C:

\[ -\frac{1}{2} + C = -\frac{1}{2} \]

\[ \text{so } C = 0 \]

Particular Solution

\[ y = \frac{-1}{\frac{1}{2}x^2 - x} \]
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Applications: Exponential Growth and Decay

\[ \frac{dy}{dt} = ky \]

This equation describes a quantity that grows or decays at a rate proportional to its size (e.g., population, interest, radioactive decay).

\( k \) is the constant of proportionality.

  • If \( k > 0 \rightarrow \) growth \( (\frac{dy}{dt} > 0) \)
  • If \( k < 0 \rightarrow \) decay \( (\frac{dy}{dt} < 0) \)

For example, for Uranium 235:

\[ k = -3.12 \times 10^{-17} \]

Newton's Law of Cooling

  • \( T(t) \): temperature as a function of time \( t \)
  • \( M \): temperature of the surrounding medium (constant)
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Newton's Law of Cooling

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\[ \frac{dT}{dt} = k(M - T) \]

Surrounding - temp: \( M \)

As long as \( M \neq T \), \( \frac{dT}{dt} \neq 0 \) until \( M = T \)

Over time, temp. of object \( \rightarrow \) temp. of surrounding.

\( k \): insulation or lack of

Solving the Differential Equation

This is separable! Let's solve it.

\( k, M \) are constants

\[ \int \frac{1}{M - T} \, dT = \int k \, dt \]
\[ -\ln |M - T| = kt + C \] \[ \ln |M - T| = -kt + C \] \[ M - T = e^{-kt} \cdot e^C \]
Try to solve for \( T \)
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\[ M - T = Ce^{-kt} \]

\[ T = M - Ce^{-kt} \]

\( (k > 0) \)

\[ \lim_{t \to \infty} e^{-kt} = 0 \]

\[ \lim_{t \to \infty} T = M \]

Another Application: Terminal Velocity

\[ \frac{dv}{dt} = g - \frac{c}{m}v \]
Also separable
\[ \frac{1}{g - \frac{c}{m}v} \, dv = dt \]