181 lines
4.3 KiB
Markdown
181 lines
4.3 KiB
Markdown
---
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date: 11.09.2024
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type: math
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---
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## Separable ODE
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- **What is it?**
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A **separable ODE** is a type of first-order differential equation where the variables can be separated on opposite sides of the equation. In other words, it can be written in the form:
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$$
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\frac{dy}{dx} = g(x)h(y),
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$$
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where the right-hand side is a product of a function of $x$ and a function of $y$. This allows the equation to be rewritten so that all $y$-terms are on one side and all $x$-terms are on the other.
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- **What are the solution steps?**
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To solve a separable ODE, follow these steps:
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1. **Rewrite the equation:** Separate the variables by moving all terms involving $y$ to one side and all terms involving $x$ to the other:
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$$
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\frac{1}{h(y)} \, dy = g(x) \, dx.
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$$
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2. **Integrate both sides:** Integrate both sides with respect to their respective variables:
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$$
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\int \frac{1}{h(y)} \, dy = \int g(x) \, dx.
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$$
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3. **Solve for $y(x)$:** Find the general solution by solving for $y$ in terms of $x$. This may involve finding an explicit or implicit form.
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4. **Apply initial conditions (if any):** If an initial condition is provided (e.g., $y(x_0) = y_0$), substitute it into the general solution to find the particular solution.
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**Example:**
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Consider the separable ODE:
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$$
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\frac{dy}{dx} = xy.
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$$
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Separating variables:
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$$
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\frac{1}{y} \, dy = x \, dx.
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$$
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Integrating both sides:
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$$
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\ln |y| = \frac{x^2}{2} + C.
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$$
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Solving for $y$, we get:
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$$
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y(x) = Ce^{x^2/2}.
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$$
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## Equidimensional (Euler–Cauchy) Equation
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An **Equidimensional (Euler–Cauchy) equation** is a type of second-order linear differential equation with variable coefficients that are powers of the independent variable $x$. It has the form:
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$$
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x^2 y'' + ax y' + b y = 0,
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$$
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where $a$ and $b$ are constants.
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- **How do we solve it?**
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To solve the Euler–Cauchy equation:
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1. **Use the substitution:** $y = x^m$, where $m$ is a constant to be determined.
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2. **Find derivatives:** Compute $y'$ and $y''$ in terms of $m$:
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$$
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y' = mx^{m-1}, \quad y'' = m(m-1)x^{m-2}.
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$$
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3. **Substitute into the original equation:** Substitute $y$, $y'$, and $y''$ into the differential equation and simplify.
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4. **Solve the characteristic equation:** The resulting equation will be a quadratic in terms of $m$:
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$$
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m(m-1) + am + b = 0.
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$$
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Solve this quadratic equation for $m$.
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5. **Form the general solution:** Depending on the roots $m_1$ and $m_2$, the general solution will be:
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- If $m_1 \neq m_2$: $y(x) = C_1 x^{m_1} + C_2 x^{m_2}$.
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- If $m_1 = m_2$: $y(x) = (C_1 + C_2 \ln x) x^{m_1}$.
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**Example:**
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Solve $x^2 y'' - 4xy' + 6y = 0$.
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1. Substitute $y = x^m$, $y' = mx^{m-1}$, and $y'' = m(m-1)x^{m-2}$.
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2. The characteristic equation becomes:
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$$
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m(m-1) - 4m + 6 = 0 \implies m^2 - 5m + 6 = 0.
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$$
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3. Solving, we find $m_1 = 2$, $m_2 = 3$.
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4. The general solution is:
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$$
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y(x) = C_1 x^2 + C_2 x^3.
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$$
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## Linear ODEs of Order 1
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A **linear ODE of order 1** is a first-order differential equation that can be written in the form:
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$$
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\frac{dy}{dx} + P(x)y = Q(x),
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$$
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where $P(x)$ and $Q(x)$ are functions of $x$.
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- **What are the rules for finding out if $\epsilon$ is homogeneous?**
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An ODE is **homogeneous** if $Q(x) = 0$. Thus, the equation becomes:
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$$
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\frac{dy}{dx} + P(x)y = 0.
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$$
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In this case, the solution involves finding an integrating factor:
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$$
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\mu(x) = e^{\int P(x) \, dx}.
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$$
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Multiplying through by $\mu(x)$ makes the left side an exact derivative:
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$$
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\frac{d}{dx} \left( \mu(x) y \right) = 0,
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$$
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which can then be integrated to solve for $y(x)$.
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## Examples of Different Types of Differential Equations
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- **Non-linear:**
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- An ODE that cannot be written in a linear form, for example:
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$$
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\frac{dy}{dx} = y^2 + x.
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$$
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The function $y^2$ makes it nonlinear.
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- **Linear, Homogeneous:**
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- An ODE where the function and its derivatives appear linearly and the right-hand side is zero:
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$$
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\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = 0.
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$$
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Here, all terms involve $y$ or its derivatives to the first power, and the equation is set to 0.
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- **Linear, Non-homogeneous:**
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- A linear ODE with a non-zero right-hand side:
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$$
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\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = e^x.
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$$
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The term $e^x$ makes it non-homogeneous. |