Lesson 4: Substitution Methods -- Bernoulli and Homogeneous

Estimated time: 35-40 minutes

Learning Objectives

Recognize Bernoulli equations and apply the substitution v = y^1-n
Recognize homogeneous equations and apply the substitution v = y/x
Reduce each type to a linear or separable equation and solve

Bernoulli Equations

Bernoulli Equation: y' + P(x)y = Q(x)yⁿ, where n ≠ 0 and n ≠ 1.

Substitution: Let v = y^1-n. Then v' = (1-n)y^-ny', and the equation becomes the linear ODE:

v' + (1-n)P(x)v = (1-n)Q(x)

Example 1: Bernoulli with n = 2

Solve y' + y = xy².

Step 1: P(x) = 1, Q(x) = x, n = 2. Let v = y^1-2 = y^-1 = 1/y.

Step 2: v' = -y^-2y'. Divide original by y²: y^-2y' + y^-1 = x.

So -v' + v = x, i.e., v' - v = -x.

Step 3: Solve the linear ODE. μ = e^-x.

d/dx[e^-xv] = -xe^-x. Integrate (by parts): e^-xv = xe^-x - e^-x + C. So v = x - 1 + Ce^x.

Step 4: Back-substitute: y = 1/v = 1/(x - 1 + Ce^x).

Homogeneous Equations

Homogeneous Equation (not to be confused with "homogeneous linear"): dy/dx = F(y/x), where the right side depends only on the ratio y/x.

Equivalently: M(x,y) dx + N(x,y) dy = 0 where M and N are homogeneous functions of the same degree.

Substitution: Let v = y/x, so y = vx and dy/dx = v + x dv/dx.

Example 2: Homogeneous Equation

Solve dy/dx = (x + y)/x = 1 + y/x.

Step 1: Let v = y/x. Then dy/dx = v + xv'.

Step 2: v + xv' = 1 + v. So xv' = 1, i.e., dv/dx = 1/x.

Step 3: Separate and integrate: v = ln|x| + C.

Step 4: Back-substitute: y/x = ln|x| + C. So y = x ln|x| + Cx.

Example 3: More Complex Homogeneous

Solve dy/dx = (x² + y²)/(2xy).

Step 1: Rewrite: F(y/x) = (1 + (y/x)²)/(2(y/x)) = (1 + v²)/(2v).

Step 2: v + xv' = (1 + v²)/(2v). So xv' = (1 + v²)/(2v) - v = (1 + v² - 2v²)/(2v) = (1 - v²)/(2v).

Step 3: Separate: 2v dv/(1-v²) = dx/x.

Integrate: -ln|1-v²| = ln|x| + C₁.

|1-v²| = A/|x| where A = e^-C₁.

Step 4: Back-substitute v = y/x: x² - y² = Ax (after simplification).

Check Your Understanding

1. For the Bernoulli equation y' + 2y = 3y³, what substitution do you use?

n = 3, so v = y^1-3 = y^-2.

2. Is dy/dx = (x² + xy)/(x²) homogeneous? If so, what substitution?

dy/dx = 1 + y/x. Yes, depends only on y/x. Substitution: v = y/x.

3. Transform y' + y/x = y²/x into a linear ODE via Bernoulli substitution.

n=2, v = 1/y, v' = -y'/y². Divide by y²: y'/y² + 1/(xy) = 1/x. So -v' + v/x = 1/x, i.e., v' - v/x = -1/x.

4. After substituting v = y/x in a homogeneous equation, what type of equation do you get?

A separable equation in v and x.

5. For n = 0 or n = 1, why is Bernoulli substitution unnecessary?

When n = 0, the equation is already linear (y' + Py = Q). When n = 1, it becomes y' + Py = Qy, which is y' + (P-Q)y = 0, a separable/linear equation.

Key Takeaways

Bernoulli: y' + Py = Qyⁿ. Substitute v = y^1-n to get a linear ODE in v.
Homogeneous: dy/dx = F(y/x). Substitute v = y/x to get a separable ODE in v.
Both methods transform a nonlinear equation into a solvable form.
Always back-substitute to express the answer in terms of the original variables.
Check for lost or singular solutions (especially y = 0 for Bernoulli).

Practice

10 problems covering all Module 2 techniques.

Practice

Module Quiz

Quiz