Separable differential equations Knowledge-Based Study

Study notes

1. What is a Separable Differential Equation?

   A first-order differential equation is said to be separable if it can be written in the form:

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

   Equivalently, dividing both sides by $h(y)$ (provided $h(y) \neq 0$) and expressing the relationship in terms of differentials yields:

   $f(y)\,dy = g(x)\,dx$

   where $f(y) = \frac{1}{h(y)}$. In this form, all expressions involving the dependent variable $y$ appear on one side with $dy$, and all expressions involving the independent variable $x$ appear on the opposite side with $dx$.

   Steps to Solve Separable Equations

   1. Separate the variables: Algebraically rearrange the equation so that every term containing $y$ multiplies $dy$ on one side, and every term containing $x$ multiplies $dx$ on the other side:
      $f(y)\,dy = g(x)\,dx$
   2. Integrate both sides: Evaluate the indefinite integrals of both expressions:
      $\int f(y)\,dy = \int g(x)\,dx$
   3. Incorporate the constant of integration: Combine the arbitrary constants into a single constant $C$ on one side, typically the side containing $x$:
      $F(y) = G(x) + C$
   4. Solve for $y$ (when possible): Obtain an explicit solution $y = y(x)$. If expressing $y$ explicitly is impractical, the solution may be left in implicit form.

   Example

   Find the general solution of the differential equation:

   $\frac{dy}{dx} = 2xy^2$

   Solution

   1. Separate variables: Assuming $y \neq 0$, divide by $y^2$ and multiply by $dx$:
      $y^{-2}\,dy = 2x\,dx$
   2. Integrate both sides:
      $\int y^{-2}\,dy = \int 2x\,dx$
   3. Compute antiderivatives:
      $-\frac{1}{y} = x^2 + C$
   4. Isolate $y$:
      $y = -\frac{1}{x^2 + C}$

   Initial Value Problems (IVPs)

   When an initial condition $y(x_0) = y_0$ is provided, substitute the known coordinates into the general solution or directly into the integrated equation to determine the unique value of the integration constant $C$.

   Singular Solutions

   When separating variables, division by $h(y)$ assumes that $h(y) \neq 0$. Any constant values $y = c$ for which $h(c) = 0$ are constant (equilibrium) solutions. If a constant solution cannot be obtained from the general family for any value of $C$, it is classified as a singular solution.

   For instance, in $\frac{dy}{dx} = 2xy^2$, setting $y = 0$ satisfies the original equation, representing an equilibrium solution that corresponds to the limit as $C \to \infty$.