Calc 2 Lecture NotesSection 7.2Page 1 of 9

Section 7.2: Separable Differential Equations

Big idea: There are many different techniques for solving diff. eqs. (Not DIFFY “Qs”). Differential equations get grouped according to their method of solution. Differential equations that can be solved by isolating the dependent and independent variables on oppositesides of the equation are called separable.

Big skill:. To be able to recognize separable differential equations, and then to solve them by separating the variables.

Separable differential equations have the form:

Their solution is given by the formula:

To solve separable differential equations:

  • Separate functions of the dependent and independent variable onto separate sides of the equation.
  • Re-write the equation so that you have differentials on both sides.
  • Integrate both sides of the equation.
  • Solve for the dependent variable, if possible.

Example: Solve the differential equation , and graph the solution(s).


Note that these are the families of hyperbolae
/

Notice that since there is an arbitrary constant of integration c, when we try to graph the solution, we should include various different graphs for various values of c. The collection of all solutions for all possible values of c is called the family of solutions.
Practice:

  1. Show that are solutions to the differential equation. Then derive that solution to the differential equation and graph the family of solutions. Also show that is an equivalent form of the solution.

  1. Solve the differential equation and graph the family of solutions. State why the solution is best left as an implicit equation. (Hint: use a conjugate…)

When the solution curve to a differential equation must pass through a specified point (x0, y0), we can re-write that condition on the curve as: y(x0) = y0. This is called an initial condition. We call the combination of a differential equation with an initial condition an initial value problem (IVP). The upshot of solving an IVP is that we must solve for a single number for the constant of integration in order to satisfy the initial condition.

Practice:

  1. Solve the initial value problem and y(0) = 1.5.

Hint: It is often easier to solve for c before you get the final form of y(x).

  1. Solve the initial value problem and y(2) = /4.

The logistic growth model for populations reflects the fact that as a population grows exponentiallyit fills up its environment, which then limits the rate of growth. If k is the (modified) growth rate ( k will have units of 1/((time)(population)) ), and M is the maximum carrying capacity of the environment, then the mathematical model is:

.

Note that y = 0 and y = M are two special solutions of this equation, corresponding to the fact that the rate of growth y = 0 when there is no population to start with, or when the population starts at the maximum carrying capacity of the environment.

Given an initial condition of (t0, y0), the solution is:
/

Proof:

Linear First-Order Differential Equationsare first-order differential equations where any occurrences of the dependent function y are linear in nature. For example, the exponential growth/decay model y(t) = –ky is linear, but the logistic growth model is not linear (due to the y2 term). Linear first order differential equations are nice because they can always be transformed into a separable differential equation. This means there is a formula for the solution to any linear first-order equation.

Derivation of the formula for the solution of a linear first-order differential equation with a constant linear coefficient:

  • Linear first-order differential equation with constant linear coefficeint:
  • Multiply both sides by :
  • Write the left hand side as a derivative of a product of functions:
  • Notice that the equation is now separable. Separate it and integrate.
  • Conclusion:
    The solution of
    is given by the formula:
  • is called anintegrating factor Because we multiply both sides by that factor so we can integrate the equation.

Practice:

  1. Solve , given the initial condition y(0) = 1.

  1. Solve , given the initial condition y(0) = 0.75.

Derivation of the formula for the solution of a general linear first-order differential equation:

  • General linear first-order differential equation:
  • Multiply both sides by an integrating factor (which we will figure out later:
  • We want the left hand side to be the derivative of a product of the functions u and y:
  • Compare what we want on the left hand side. By equating coefficients, get an equation for u(t), and solve that equation:
  • Thus the solution of the original differential equation is:
  • u(t) is also called anintegrating factor.

Practice:

  1. Solve using this technique, given the initial condition y(0) = 0.