Introduction to Differential Equations and Numerical Methods

Verifying Solutions to Differential Equations

• A differential equation involves a function $y$ and one or more of its derivatives. To check if a given expression is a solution, one must determine the necessary derivatives and substitute them into the original equation.

• Example Verification:

  • Given differential equation: $y'' - 3y' + 2y = 24$
  • Possible solution: $y = 3e^x + 4e^{2x} + 2e^{-2x}$
  • First derivative ($y'$):
    • Using the chain rule: $\frac{d}{dx}(3e^x) = 3e^x$
    • $\frac{d}{dx}(4e^{2x}) = 8e^{2x}$
    • $\frac{d}{dx}(2e^{-2x}) = -4e^{-2x}$
    • Result: $y' = 3e^x + 8e^{2x} - 4e^{-2x}$
  • Second derivative ($y''$):
    • $\frac{d}{dx}(3e^x) = 3e^x$
    • $\frac{d}{dx}(8e^{2x}) = 16e^{2x}$
    • $\frac{d}{dx}(-4e^{-2x}) = 8e^{-2x}$
    • Result: $y'' = 3e^x + 16e^{2x} + 8e^{-2x}$
  • Substitution and Simplification:
    • Term 1 ($y''$): $3e^x + 16e^{2x} + 8e^{-2x}$
    • Term 2 ($-3y'$): $-3(3e^x + 8e^{2x} - 4e^{-2x}) = -9e^x - 24e^{2x} + 12e^{-2x}$
    • Term 3 ($2y$): $2(3e^x + 4e^{2x} + 2e^{-2x}) = 6e^x + 8e^{2x} + 4e^{-2x}$
  • Summation of exponents:
    • $e^x$ terms: $3 - 9 + 6 = 0$
    • $e^{2x}$ terms: $16 - 24 + 8 = 0$
    • $e^{-2x}$ terms: $8 + 12 + 4 = 24$
  • Result: $24 = 24$. This expression is indeed a solution to the differential equation.

The Order of Differential Equations

• The order of a differential equation is defined as the highest order of derivative that appears in the equation. This is analogous to the degree of a polynomial, which is the highest power found within it.

• Examples of Order:

  • $y' = 2 - x$: This contains only a first derivative, making it a first-order differential equation.
  • $y'' - 3y' + 2y = 24$: The highest derivative is the second derivative, making it a second-order differential equation.
  • An equation containing $y'''$ is a third-order differential equation.

• Notation for Higher Orders:

  • Beyond high orders (typically after the third derivative), the use of apostrophes/primes becomes impractical (e.g., writing ten apostrophes for the tenth derivative).
  • Instead, higher derivatives are written with a superscript in parentheses, such as $y^{(4)}$ for the fourth derivative or $y^{(10)}$ for the tenth derivative.
  • Context is usually sufficient to distinguish these from powers (e.g., $y^4$). In differential equations, one would not typically take a negative derivative or a fractional derivative (like a half-derivative) in this context.

General vs. Particular Solutions

• A general solution contains an arbitrary constant, usually denoted as $C$. This constant represents the fact that many different functions can satisfy the same relationship between derivatives, because the derivative of a constant is zero.

• A particular solution is a specific function that solves the equation and satisfies a given set of conditions, where a specific value for $C$ has been determined.

• Examples:

  • For the equation $y' = 2x$, the solution $y = x^2$ is a solution, but so is $y = x^2 + 1$.
  • The general solution is $y = x^2 + C$, where the value of $C$ is not yet chosen.
  • A particular solution might be $y = x^2 + 1$.

Initial Value Problems (IVP)

• When a differential equation is accompanied by a condition, such as a requirement that the solution function $y = f(x)$ passes through a specific point $(x, y)$, it is called an initial value problem.

• Initial conditions help model real-world situations where the starting state is known (e.g., position at time $t=0$, temperature at start time).

• Example: Solving $y' = 2 - x$ through point $(2, 7)$

  • Step 1 (General Solution): Integrate both sides to find the antiderivative: $y = 2x - \frac{x^2}{2} + C$.
  • Step 2 (Particular Solution): Substitute the point inputs ($x=2$, $y=7$) to find $C$.
  • $7 = 2(2) - \frac{(2)^2}{2} + C$
  • $7 = 4 - 2 + C$
  • $7 = 2 + C \rightarrow C = 5$
  • Particular Solution: $y = 2x - \frac{x^2}{2} + 5$.

• Example: Anti-derivative for $y' = 4x + 3$ passing through $(1, 7)$

  • Antiderivative: $y = 2x^2 + 3x + C$.
  • Substitute values: $7 = 2(1)^2 + 3(1) + C$.
  • $7 = 2 + 3 + C$
  • $7 = 5 + C \rightarrow C = 2$
  • Particular Solution: $y = 2x^2 + 3x + 2$.

• Further Verification Example:

  • Equation: $y' + 2y = e^t$ with initial condition $y(0) = 3$.
  • Possible solution: $y = 2e^{-2t} + e^t$.
  • Check condition: $y(0) = 2e^0 + e^0 = 2(1) + 1 = 3$. Satisfied.
  • Check derivative: $y' = -4e^{-2t} + e^t$.
  • Plug into equation: $(-4e^{-2t} + e^t) + 2(2e^{-2t} + e^t)$.
  • $-4e^{-2t} + e^t + 4e^{-2t} + 2e^t = 3e^t$. (Note: There was a adjustment in calculation during the session; the goal was to verify if the function satisfies the specific differential relationships provided).

Physics Application: Gravity and Motion

• This modeling utilizes the relationships between acceleration, velocity, and position via derivatives and integrals.

  • Velocity $v(t)$ is the derivative of position $s(t)$. ($s'(t) = v(t)$).
  • Acceleration $a(t)$ is the derivative of velocity $v(t)$. ($v'(t) = a(t)$).

• Scenario: A baseball ($m = 0.15\,kg$) is thrown upward from a height of $3\,m$ with an initial velocity of $10\,m/s$. Gravity is the only force ($a = -9.8\,m/s^2$).

• Step 1: Finding Velocity $v(t)$

  • $v'(t) = -9.8$
  • Antiderivative: $v(t) = -9.8t + C$.
  • Initial velocity $v(0) = 10 \rightarrow 10 = -9.8(0) + C \rightarrow C = 10$.
  • Velocity function: $v(t) = -9.8t + 10$.
  • Velocity at $t=2$: $v(2) = -9.8(2) + 10 = -19.6 + 10 = -9.6\,m/s$. (The ball is moving downward).

• Step 2: Finding Position $s(t)$

  • $s'(t) = -9.8t + 10$
  • Antiderivative: $s(t) = -4.9t^2 + 10t + C$.
  • Initial height $s(0) = 3 \rightarrow 3 = -4.9(0) + 10(0) + C \rightarrow C = 3$.
  • Position function: $s(t) = -4.9t^2 + 10t + 3$.

• Note: For this calculation, the mass ($0.15\,kg$) of the ball is irrelevant to finding the position and velocity in a vacuum (gravity-only) scenario.

Direction Fields (Slope Fields)

• When an explicit solution is difficult to find, numerical or graphical methods provide insight. A direction field is a graph consisting of small arrows representing the slope of the tangent line ($y'$) at various points $(x, y)$.

• Mechanism:

  • The output of the differential equation $y' = f(x, y)$ describes the slope of the tangent line at that point.
  • Arrows track where the curve is moving. If a curve passes through a point, it must follow the direction of the arrow at that point.

• Dependences:

  • If the equation involves only $y$ (e.g., $y' = 0.4y(1 - \frac{y}{72})$), the slopes are consistent across all $x$ values for a fixed $y$. These are often seen in fluid dynamics or population modeling.
  • If the equation involves both $x$ and $y$ (e.g., $y' = 3x + 2y - 4$), every point $(x, y)$ must be calculated individually.

• Equilibrium Solutions:

  • An equilibrium solution occurs when the derivative $y'$ is zero for all $x$, resulting in a constant horizontal line solution.
  • Example: $y' = (x - 3)(y - 2)(y + 2)$.
  • Setting $y' = 0$ yields horizontal solutions at $y = 2$ and $y = -2$.
  • While $x = 3$ also makes the derivative zero, it represents a vertical line, which is not a function and therefore not a valid solution for $y(x)$.

Euler's Method (Numerical Approximation)

• Euler's Method uses linear approximations to estimate points on a solution curve when an antiderivative cannot be found.

• Formula for Linearization: $L(x) = f'(a)(x - a) + f(a)$.

• Example Process ($y' = x^2 - y^2$ starting at $(-1, 2)$):

  • 1. Calculate the initial slope: $y'(-1, 2) = (-1)^2 - (2)^2 = 1 - 4 = -3$.
  • 2. Define the linearization function at this point: $L(x) = -3(x - (-1)) + 2 = -3(x + 1) + 2$.
  • 3. Take a small step ($h$ or $\Delta x$) to a new $x$ value, e.g., from $x = -1$ to $x = -0.9$.
  • 4. Estimate the new $y$ value: $y(-0.9) \approx -3(-0.9 + 1) + 2 = -3(0.1) + 2 = 1.7$.
  • 5. Use the new point $(-0.9, 1.7)$ to find a new slope for the next step:
    • $y' = (-0.9)^2 - (1.7)^2 = 0.81 - 2.89 = -2.08$.
  • 6. Rebuild the linearization: $L_2(x) = -2.08(x - (-0.9)) + 1.7$.
  • 7. Continue this process iteratively to follow the curve.

• Critical Observations:

  • The accuracy of Euler's method depends on the step size. Smaller steps lead to better estimations but more calculations.
  • It is an estimation off of an underestimation/overestimation, meaning it can accumulate error as you move further from the starting point.

Questions & Discussion

• Question: How do you make the notation for the fourth derivative unique from a power?

• Response: In the context of differential equations, parentheses are used around the superscript, like $y^{(4)}$. Furthermore, context usually clarifies it because it is unlikely a specific power would be present without indicating a derivative in this curriculum. One wouldn't typically have something like a "half-derivative" in this basic context either.

• Question: Is the initial point irrelevant in the slope field verification problem?

• Response: No, the initial condition was satisfied earlier in that problem ($y(0) = 3$). Often, constants are absorbed into other functions during integration, so the point remains necessary to ensure the solution is particular to the given scenario.