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Solution to $ y' = ay + b $ (Continued)

Last time:

\[ \frac{dy}{dt} = -2y + 5 \]

Solution: $ y(t) = \frac{5}{2} + C e^{-2t} $

“general solution”

$ \rightarrow $ infinitely-many curves, depending on $ C $

Concept Note

The constant $ C $ is an arbitrary constant of integration. Varying $ C $ generates a one-parameter family of integral curves.

To find $ C $, normally we need an initial condition $ y(0) = y_0 $

$ \rightarrow $ this is now an initial-value problem (IVP)

Example

For example, suppose $ y(0) = 1 $ for the differential equation $ y' = -2y + 5 $:

\[ y(t) = \frac{5}{2} + C e^{-2t} \]

\[ y(0) = 1 = \frac{5}{2} + C e^0 = \frac{5}{2} + C \implies C = -\frac{3}{2} \]

So, \[ y(t) = \frac{5}{2} - \frac{3}{2} e^{-2t} \]

“particular solution”
(one out of the general)

Key Distinction

A general solution contains arbitrary constants, whereas a particular solution is fixed to a single curve passing through the specified initial state.

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Physical Interpretation: Free Fall with Air Resistance

Let’s interpret $ \frac{dy}{dt} = -2y + 5 $ in the context of free fall:

Free-body diagram of a falling body showing upward drag force c v and downward gravity m g — Figure: Free-body diagram of a falling body showing upward drag force $ c v $ and downward gravity $ m g $ Show Details

Free-body diagram illustrating the forces acting on a falling object: downward gravitational force $ m g $ and opposing upward drag/resistance force $ c v $.

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

$ \frac{dy}{dt} = -2y + 5 $ is equivalent to:

$ g = 5 $
$ \frac{c}{m} = 2 $ (resistance due to air is twice the mass)

Behavior of the Particular Solution

\[ y(t) = \frac{5}{2} - \frac{3}{2} e^{-2t}, \quad y(0) = 1 \]

Graph of the solution curve y(t) starting at (0, 1) and asymptotically approaching y = \frac{5}{2} — Figure: Graph of the solution curve $ y(t) $ starting at $ (0, 1) $ and asymptotically approaching $ y = \frac{5}{2} $ Show Details

Coordinate plot showing the solution trajectory on the $ (t, y) $ plane. Starting at the initial value $ y(0) = 1 $, the curve asymptotically increases towards the horizontal asymptote at $ y = \frac{5}{2} $ as time $ t \to \infty $.

As $ t \to \infty $, $ e^{-2t} \to 0 $

So $ y \to \frac{5}{2} $

Terminal Velocity

As $ t \to \infty $, the exponential decay term $ e^{-2t} $ approaches zero, meaning the velocity $ y(t) $ converges to its equilibrium (terminal) velocity $ v_{\text{term}} = \frac{5}{2} $.

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Terminal Velocity & Differential Equation Analysis

In this case, we started w/ initial velocity of $ 1 $. Then as time goes on, we pick up more velocity and eventually approach $ \frac{5}{2} $ but never more $ \to $ terminal velocity.

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

Constant battle between $ g $ (gravity) and $ \frac{c}{m}v $ (drag)

Physical Meaning: The parameter $ g $ represents acceleration due to gravity, while $ -\frac{c}{m}v $ represents viscous air resistance (drag) opposing the direction of motion, proportional to velocity.

Direction Field & Equilibrium Velocity

Figure: direction field and phase curve plot for velocity approaching terminal velocity v = \frac{mg}{c} — Figure: Figure: direction field and phase curve plot for velocity approaching terminal velocity $ v = \frac{mg}{c} $ Show Details

Coordinate graph showing velocity $ v(t) $ versus time $ t $. The horizontal dashed line represents the stable equilibrium velocity $ v = \frac{mg}{c} $ where $ v' = 0 $. In the upper region where drag dominates ($ v > \frac{mg}{c} $), slope field vectors point downward causing solutions to decrease asymptotically toward $ \frac{mg}{c} $. In the lower region where weight dominates ($ v < \frac{mg}{c} $), slope field vectors point upward causing solutions to increase asymptotically toward $ \frac{mg}{c} $.

Equilibrium Condition:

\[ v' = 0 \implies v = \frac{mg}{c} \]

Drag dominates: When $ v > \frac{mg}{c} $, $ v' < 0 $, so velocity decreases toward $ \frac{mg}{c} $.
Weight dominates: When $ v < \frac{mg}{c} $, $ v'> 0 $, so velocity increases toward $ \frac{mg}{c} $.

Population Dynamics Interpretation

Another interpretation of $ \frac{dy}{dt} = ay + b $:

In terms of population, rate is proportional to $ a $ times size plus a constant addition ($ b > 0 $) at all times.

Concept Breakdown: Here, $ ay $ represents endogenous exponential population growth or decline, while $ b $ represents a continuous constant immigration rate (if $ b > 0 $) or emigration/harvesting rate (if $ b < 0 $).

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Classification of Diff. Eqs.

Two main types:

Ordinary Diff. Eq. (ODE)

Only contain ordinary derivatives
Solution is single-variable

Partial Diff. Eq. (PDE)

Can contain partial derivatives
Solution is multivariate

In MA 36600, we study ODE (in MA 30300 there is PDE).

Some Example ODEs:

List of example ordinary differential equations, their physical laws, and solution variables
Differential Equation	Physical Law / System	Solution Form
$ \frac{dv}{dt} = g - \frac{c}{m}v $	Falling body with air resistance	$ v(t) $
$ \frac{dT}{dt} = k(T - T_{\text{env}}) $	Newton's Law of Cooling	$ T(t) $

Newton's Law of Cooling: States that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body $ T(t) $ and its surrounding environment $ T_{\text{env}} $, where $ k $ is a thermal transfer constant.

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Differential Equations: Order and Linearity

Classification by Type, Order, and Linearity

Example Partial Differential Equation (PDE)

Wave Equation

$$a^2 \frac{\partial^2 y(x,t)}{\partial x^2} = \frac{\partial^2 y(x,t)}{\partial t^2}$$

Solution: $ y(x,t) $

A partial differential equation (PDE) contains partial derivatives with respect to multiple independent variables (here, position $ x $ and time $ t $).

Order of a Differential Equation

Order of a diff. eq.: order of the highest derivative

Examples of Differential Equations and their respective Orders
Differential Equation	Type & Order
$$\frac{dv}{dt} = g - \frac{c}{m} v$$	1st order ODE
$$m y'' + c y' + k y = 0$$	2nd order ODE
$$a^2 \frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial t^2}$$	2nd order PDE

Linear vs. Nonlinear

• In linear eqs., the coefficients of the dependent variable and its derivatives (e.g., $ y, y', y'', \text{etc.} $) do not contain the dependent variable and its derivatives.

Linear equations only allow dependent variables and their derivatives to appear to the first power, without products or nonlinear functions (such as $ \sin y $ or $ e^y $) of them.

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Linear vs. Nonlinear: Illustrative Examples

Step-by-step coefficient analysis for testing linearity

Linear Differential Equations

Example 1

$$x^2 y'' + (\cos x) y' + 5 y = e^x$$

Do the coefficients $ x^2 $, $ \cos x $, $ 5 $, and the right-hand side $ e^x $ contain $ y, y', y'', \text{etc.} $?

NO. So this eq. is linear.

Example 2

$$y''' + (\sin x) y = e^{x^2} \tan x$$

Linear

Nonlinear Differential Equations

Example 3

$$y'' + (y')^2 + y = 0$$

$$y'' + \underbrace{(y')}_{\text{coeff. } y' \text{ contains } y'} y' + y = 0 \quad \implies \quad \textbf{nonlinear}$$

Because $ (y')^2 = (y') \cdot y' $, the effective coefficient multiplying the derivative $ y' $ is $ y' $ itself, which violates linearity.

Example 4

$$y'' + \underbrace{\sin y}_{\text{this makes the eq. nonlinear}} = 0$$

$$y'' = -\sin y$$

(Like $ e^x $ in the 1st example, but here the function contains the dependent variable $ y $.)

Therefore, this equation is nonlinear.

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Solutions to Differential Equations

Solution:

A function that satisfies the differential equation.

For example, $ y = e^t $ is a solution to $ y' = y $

(so is $ 2e^t $, $ 3e^t $, etc.)

For example, for the equation $ y'' = y $:

$ y = e^t $ and $ y = e^{-t} $ are two possible solutions.

For the equation $ y'' = -y $:

$ y = \cos x $ is a solution; so is $ y = \sin x $.

(Part of this course is on techniques to find the solutions systematically.)

Homogeneous vs. Nonhomogeneous

If written in the form of

\[ F(y, y', y'', y''', \text{etc.}) = g(x) \]

where $ g(x) = 0 $, then the equation is homogeneous.

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Homogeneous vs. Nonhomogeneous Examples

\[ \underbrace{y'' + 2y' - 3y}_{F(y, y', y'')} = \underbrace{\cos x}_{\text{Is this } 0\text{?}} \]

If $ g(x) = 0 $:

The equation is homogeneous.

If $ g(x) \neq 0 $:

The equation is nonhomogeneous.

Since $ \cos x \neq 0 $, the equation $ y'' + 2y' - 3y = \cos x $ is nonhomogeneous.

\[ y'' + 2y' - 3y = 0 \quad \text{is homogeneous} \]

Solution to \( y' = ay + b \) (Continued)