Lesson 4 (08/31/16)

Lesson 4 (08/31/16)

Seperable equations and Applications (1.4)

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Office Hours: MWF 1:30 – 2:30, MATH 842

Review Examples

1. Find equilibrium solution to \( \frac{dy}{dx} = x(y - 4) \)

\( y' = f(x, y) \)
want \( f(x, y) = 0 \) + satisfies \( \frac{dy}{dx} = 0 \)

\[ x(y - 4) = 0 \] \[ x = 0 \quad \text{or} \quad y = 4 \]

For \( y = 4 \), \( \frac{dy}{dx} = 0 \).

\( y = 4 \) is the only equilibrium solution

2. Which of the following IVP has a Unique Solution?

Condition for Uniqueness:
For \( y' = f(x, y) \), \( y(x_0) = y_0 \), if \( f(x, y) \) and \( f_y \) are both continuous at \( (x_0, y_0) \).

I. (Incorrect / No unique solution guaranteed)

\[ y' = \sqrt{|y|}, \quad y(0) = 0 \]

\( \frac{\partial}{\partial y} \sqrt{|y|} \) not continuous at \( (0, 0) \)

II. (Incorrect / No unique solution guaranteed)

\[ y' = \frac{1}{x}, \quad y(0) = 0 \]

\( \frac{1}{x} \) is not continuous at \( (0, 0) \)

III. (Correct / Has a unique solution)

\[ y' = \frac{1}{x}, \quad y(1) = 7 \]

both \( f(x, y) = \frac{1}{x} \), \( f_y = 0 \) are continuous at \( (1, 7) \)

IV. (Correct / Has a unique solution)

\[ y' = -3y, \quad y(0) = 5 \]

\( f(x, y) = -3y \), \( f_y = -3 \) are both continuous at \( (0, 5) \)

Finding solutions to \( \frac{dy}{dx} = \frac{f(x)}{g(y)} \)

Review Chain Rule

\[ \frac{d}{dx} \left[ \sin(x^3) \right] = \cos(x^3) \cdot 3x^2 \]

\( y \) is a function of \( x \)

\[ \frac{d}{dx} \left[ G(y) \right] = G'(y) \cdot \frac{dy}{dx} \]

Suppose \( G(y) \) is an antiderivative of \( g(y) \)

\[ \int \frac{d}{dx} G(y) \, dx = \int g(y) \frac{dy}{dx} \, dx \] \[ \int g(y) \frac{dy}{dx} \, dx = G(y) \]

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

idea! take \(g(y)\) to the left

OR

seperate \(x\) & \(y\) variables

then integrate on both sides

\[ \int \left[ g(y) \frac{dy}{dx} \right] dx = \int f(x) \, dx + C \]

OR

\[ \int g(y) \, dy = \int f(x) \, dx + C \]

eg:

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

divide with \( 1+y^2 \)

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

integrate both sides

"Can only do this when \( 1+y^2 \neq 0 \)" ✓

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

Solution to \( y' = x(1+y^2) \)

is implicitly given by \( \tan^{-1}(y) = \frac{x^2}{2} + C \)

OR

Solve for \( y \) as \( y = \tan\left(\frac{x^2}{2} + C\right) \)

Example:

\[ \frac{dy}{dx} = x(y-4) \]

\(y \neq 4 \rightarrow\)

\[ \frac{1}{y-4} \, dy = x \, dx \] \[ \int \frac{1}{y-4} \, dy = \int x \, dx + C \] \[ \ln |y-4| = \frac{x^2}{2} + C \quad \text{Implicit solution} \]

We know \(y=4\) is a solution.

Is it possible to find \(C\) such that \(y=4\) is solution?

NO!!

\[ \ln |y-4| = \frac{x^2}{2} + C \quad \text{is General solution} \]

ignoring \(y=4\) solution

Singular solution

Example:

\[ \frac{dy}{dx} = 6x (y-1)^{2/3} \]

\[ y(x) = 1 \leadsto \frac{dy}{dx} = 0 = 6x(1-1)^{2/3} \implies y(x) = 1 \text{ is equilibrium solution} \]

Also singular bcz we seperate variables as long as \( y \neq 1 \)

Case \( y \neq 1 \):

\[ \frac{1}{(y-1)^{2/3}} \, dy = 6x \, dx \]

\[ \int \frac{1}{(y-1)^{2/3}} \, dy = \int 6x \, dx + c \]

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

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

\[ y(x) = (x^2 + k)^3 + 1 \]

General solution except the solution \( y(x) = 1 \)

If we try to find \(k\) such that \(y(x) = 1\)

\[ x = (x^2 + k)^3 + x \] \[ (x^2 + k)^3 = 0 \]

\(k\) is supposed to be a const. but if we solve we get

\[ k = -x^2 \]

Not constant

\(y = 1\) is a Singular Solution

Example: Find Solution to IVP

Eg: Find solution to IVP \(\frac{dy}{dx} = e^{-y}(2x-4)\), \(y(5)=0\)

multiply with \(e^y \neq 0\) on both sides

\[e^y dy = (2x-4) dx \leadsto \int e^y dy = \int 2x-4 \, dx + C\] \[e^y = x^2 - 4x + C\]

OR

\[y(x) = \ln(x^2 - 4x + C)\]

IVP: \(y(5)=0 \leadsto 0 = y(5) = \ln(25-20+C)\)

\[0 = \ln(5+C)\] \[5+C = 1 \implies C = -4\]

Solution to IVP: \(y(x) = \ln(x^2-4x-4)\)

Some Applications where Model is a Separable Equation

1. Exponential growth / decay:

\[ p'(t) = k p(t) \]

\( k > 0 \text{ is growth} \)
\( k < 0 \text{ is decay} \)

Solution: \( p(t) = A e^{kt} \) (Verify!!)

2. Newton's Law of Cooling

\(T(t) =\) Temperature of an object at time \(t\)

\(T_0 =\) Constant temperature of Medium (Room temp.)

\[ T'(t) = k(T_0 - T) \]

\(k\) is Constant.

3. Logistic Equation

\( P(t) = \) population of a quantity

\( r = \) rate of growth

\( K = \) carrying capacity

\[ P'(t) = rP\left(1 - \frac{P}{K}\right) \]

Review integration!

eg: \[ \int \frac{1}{y(y-1)} \, dy \quad ? \]