Notes/Calculus 2/Solving 1st order ODE.md

date	type
11.09.2024	math

Separable ODE

• What is it?

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:

\frac{dy}{dx} = g(x)h(y),

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.

• What are the solution steps?

To solve a separable ODE, follow these steps:

1. Rewrite the equation: Separate the variables by moving all terms involving y to one side and all terms involving x to the other:

   
   \frac{1}{h(y)} \, dy = g(x) \, dx.
   

2. Integrate both sides: Integrate both sides with respect to their respective variables:

   
   \int \frac{1}{h(y)} \, dy = \int g(x) \, dx.
   

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.

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.

Example: Consider the separable ODE:

\frac{dy}{dx} = xy.

Separating variables:

\frac{1}{y} \, dy = x \, dx.

Integrating both sides:

\ln |y| = \frac{x^2}{2} + C.

Solving for y, we get:

y(x) = Ce^{x^2/2}.

Equidimensional (Euler–Cauchy) Equation

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:

x^2 y'' + ax y' + b y = 0,

where a and b are constants.

• How do we solve it?

To solve the Euler–Cauchy equation:

1. Use the substitution: y = x^m, where m is a constant to be determined.

2. Find derivatives: Compute y' and y'' in terms of m:

   
   y' = mx^{m-1}, \quad y'' = m(m-1)x^{m-2}.
   

3. Substitute into the original equation: Substitute y, y', and y'' into the differential equation and simplify.

4. Solve the characteristic equation: The resulting equation will be a quadratic in terms of m:

   
   m(m-1) + am + b = 0.
   

   Solve this quadratic equation for m.

5. Form the general solution: Depending on the roots m_1 and m_2, the general solution will be:

   • If m_1 \neq m_2: y(x) = C_1 x^{m_1} + C_2 x^{m_2}.
   • If m_1 = m_2: y(x) = (C_1 + C_2 \ln x) x^{m_1}.

Example: Solve x^2 y'' - 4xy' + 6y = 0.

1. Substitute y = x^m, y' = mx^{m-1}, and y'' = m(m-1)x^{m-2}.

2. The characteristic equation becomes:

   
   m(m-1) - 4m + 6 = 0 \implies m^2 - 5m + 6 = 0.
   

3. Solving, we find m_1 = 2, m_2 = 3.

4. The general solution is:

   
   y(x) = C_1 x^2 + C_2 x^3.
   

Linear ODEs of Order 1

A linear ODE of order 1 is a first-order differential equation that can be written in the form:

\frac{dy}{dx} + P(x)y = Q(x),

where P(x) and Q(x) are functions of x.

• What are the rules for finding out if \epsilon is homogeneous?

An ODE is homogeneous if Q(x) = 0. Thus, the equation becomes:

\frac{dy}{dx} + P(x)y = 0.

In this case, the solution involves finding an integrating factor:

\mu(x) = e^{\int P(x) \, dx}.

Multiplying through by \mu(x) makes the left side an exact derivative:

\frac{d}{dx} \left( \mu(x) y \right) = 0,

which can then be integrated to solve for y(x).

Examples of Different Types of Differential Equations

• Non-linear:

  • An ODE that cannot be written in a linear form, for example:

  
  \frac{dy}{dx} = y^2 + x.
  

  The function y^2 makes it nonlinear.

• Linear, Homogeneous:

  • An ODE where the function and its derivatives appear linearly and the right-hand side is zero:

  
  \frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = 0.
  

  Here, all terms involve y or its derivatives to the first power, and the equation is set to 0.

• Linear, Non-homogeneous:

  • A linear ODE with a non-zero right-hand side:

  
  \frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = e^x.
  

  The term e^x makes it non-homogeneous.