2.1 Separable Equations

Separable equations are a key type of first-order differential equations. They're special because we can split the variables, making them easier to solve. This technique is super useful and pops up in many real-world problems.

Solving separable equations involves integrating both sides after splitting the variables. We get a general solution with a constant C, which we can fine-tune using initial conditions to find particular solutions.

Separable Differential Equations

Definition and Characteristics

• Separable differential equation can be written in the form $\frac{dy}{dx} = f(x)g(y)$
  • $f(x)$ is a function of $x$ only
  • $g(y)$ is a function of $y$ only
• Separable equations allow variables $x$ and $y$ to be separated on opposite sides of the equation
  • Enables solving the equation by integrating each side independently
• Separation of variables technique used to solve separable differential equations
  • Rearrange the equation to have all terms involving $x$ on one side and all terms involving $y$ on the other side
  • Divide both sides by the coefficient of $dy$ to isolate $dy$ on the left-hand side
    • $\frac{dy}{g(y)} = f(x)dx$

Integration and Solution

• Integrate both sides of the separated equation with respect to their respective variables
  • $\int \frac{dy}{g(y)} = \int f(x)dx$
• Integration yields an equation relating $x$ and $y$, which is the general solution to the separable differential equation
  • Solve for $y$ explicitly if possible to obtain an explicit solution
  • Leave the solution in implicit form if explicit solution is not easily obtainable
• Add a constant of integration $(C)$ to the right-hand side after integrating
  • Represents an arbitrary constant introduced during the integration process
  • Determined by initial conditions or boundary conditions of the problem

Solutions to Separable Equations

General and Particular Solutions

• General solution to a separable differential equation obtained by integrating both sides
  • Contains an arbitrary constant of integration $(C)$
  • Represents a family of solutions that satisfy the differential equation
  • Example: $\int \frac{dy}{y} = \int xdx \implies \ln|y| = \frac{1}{2}x^2 + C$
• Particular solution derived from the general solution by specifying a value for the constant of integration
  • Obtained by using additional information or conditions given in the problem
  • Represents a specific solution curve that satisfies both the differential equation and the given condition
  • Example: If $y(0) = 1$, then $C = 0$ and the particular solution is $\ln|y| = \frac{1}{2}x^2$

Implicit and Explicit Solutions

• Implicit solution is an equation relating $x$ and $y$ without explicitly solving for $y$
  • Obtained directly from the integration step in solving separable equations
  • May be more convenient to leave the solution in implicit form if explicit solution is complex or not easily obtainable
  • Example: $\int \frac{dy}{y^2+1} = \int dx \implies \arctan(y) = x + C$
• Explicit solution is an equation where $y$ is expressed explicitly as a function of $x$
  • Obtained by solving the implicit solution for $y$ in terms of $x$
  • Provides a direct relationship between $y$ and $x$
  • Example: $\arctan(y) = x + C \implies y = \tan(x + C)$

Initial Value Problems

Definition and Solution Process

• Initial value problem (IVP) is a differential equation along with an initial condition specifying the value of the dependent variable at a particular point
  • Differential equation describes the rate of change of a function
  • Initial condition provides a specific starting point for the solution
  • Example: $\frac{dy}{dx} = xy$, $y(0) = 1$
• Solving an IVP for a separable equation involves:
  1. Separating variables and integrating both sides of the equation
  2. Applying the initial condition to determine the value of the constant of integration $(C)$
  3. Substituting the value of $C$ into the general solution to obtain the particular solution satisfying the IVP

Determining the Constant of Integration

• Initial condition provides a known point $(x_0, y_0)$ that the solution must pass through
  • Substitute the values of $x_0$ and $y_0$ into the general solution
  • Solve the resulting equation for the constant of integration $(C)$
• Example: For the IVP $\frac{dy}{dx} = xy$, $y(0) = 1$, the general solution is $\ln|y| = \frac{1}{2}x^2 + C$
  • Substitute $(0, 1)$ into the general solution: $\ln|1| = \frac{1}{2}(0)^2 + C \implies C = 0$
  • Particular solution satisfying the IVP: $\ln|y| = \frac{1}{2}x^2$