ST13 — Student Version

Overview

This tutorial covers Bernoulli differential equations, including identification, transformation to linear form, and solution methods. It also includes problems requiring multiple solution approaches for the same differential equation.

How to Solve a Bernoulli Differential Equation

What is a Bernoulli DE?

A Bernoulli differential equation is any first-order ODE that can be written in the standard form:

$\frac{dy}{dx} + P(x)y = Q(x)y^n$

where $P(x)$ and $Q(x)$ are functions of $x$ only, and $n$ is any real number except 0 and 1.

If $n = 0$: It's just a linear first-order ODE.
If $n = 1$: It's a separable equation.
If $n \neq 0, 1$: It's Bernoulli and requires a substitution.

The General Method

Step 1: Rewrite in standard form Get the equation into the form: $\frac{dy}{dx} + P(x)y = Q(x)y^n$

Step 2: Identify $n$ Find the exponent $n$ on the right-hand side.

Step 3: Divide by $y^n$ Divide every term by $y^n$ (assuming $y \neq 0$): $y^{-n}\frac{dy}{dx} + P(x)y^{1-n} = Q(x)$

Step 4: Make the substitution Let $v = y^{1-n}$.

Then differentiate $v$ with respect to $x$: $\frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}$

So: $y^{-n}\frac{dy}{dx} = \frac{1}{1-n}\frac{dv}{dx}$

Step 5: Transform to linear form Substitute into the equation from Step 3: $\frac{1}{1-n}\frac{dv}{dx} + P(x)v = Q(x)$

Multiply by $(1-n)$: $\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$

This is now a linear first-order ODE in $v$.

Step 6: Solve the linear equation Use the integrating factor method:

Find $\mu(x) = e^{\int (1-n)P(x),dx}$
Multiply through by $\mu(x)$
Left side becomes $\frac{d}{dx}[\mu(x) v]$
Integrate both sides: $\mu(x) v = \int \mu(x)(1-n)Q(x),dx + C$
Solve for $v$

Step 7: Back-substitute Replace $v = y^{1-n}$ to get the solution in terms of $y$.

Worked Example

Problem: Solve $\frac{dy}{dx} - y = e^x y^2$

Step 1: Already in standard form with $P(x) = -1$ and $Q(x) = e^x$.

Step 2: The exponent on the right is $n = 2$.

Step 3: Divide by $y^2$: $y^{-2}\frac{dy}{dx} - y^{-1} = e^x$

Step 4: Substitute $v = y^{1-2} = y^{-1} = \frac{1}{y}$

Differentiate: $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$

So: $y^{-2}\frac{dy}{dx} = -\frac{dv}{dx}$

Step 5: Substitute: $-\frac{dv}{dx} - v = e^x$

Multiply by $-1$: $\frac{dv}{dx} + v = -e^x$

Step 6: Integrating factor: $\mu(x) = e^{\int 1,dx} = e^x$

Multiply through: $e^x\frac{dv}{dx} + e^x v = -e^{2x}$

Left side is $\frac{d}{dx}[e^x v]$: $\frac{d}{dx}[e^x v] = -e^{2x}$

Integrate: $e^x v = -\frac{1}{2}e^{2x} + C$

Solve for $v$: $v = -\frac{1}{2}e^x + Ce^{-x}$

Step 7: Back-substitute $v = \frac{1}{y}$: $\frac{1}{y} = -\frac{1}{2}e^x + Ce^{-x}$

Or: $y = \frac{1}{Ce^{-x} - \frac{1}{2}e^x}$

Questions Summary

Bernoulli Equations: Show that each differential equation is Bernoulli and solve it.
- (a) $x\frac{dy}{dx} + y = \frac{1}{y^2}$
- (b) $\frac{dy}{dx} - y = e^x y^2$
- (c) $xe^y y' = 2e^y + x^3 e^{2y}$
- (d) $x^2 y' + 2xy = 5y^3$
- (e) $x^2 \frac{dy}{dx} + y^2 \cos x = 3xy$
- (f) $3(1 + x^2)y' = 2xy(y^3 - 1)$
Particular Solutions: Obtain particular solutions for Bernoulli equations with initial conditions.
- (a) $xdy + ydx = x^3y^6$; $y(1) = 1$
- (b) $2xyy' = y^2 - 2x^3$; $y(2) = 1$
- (c) $(y^4 - 2xy)dx + 3x^2 dy = 0$; $y(2) = 1$
- (d) $(x^2 + 2y^2)dx - ydy = 0$; $y(0) = 1$
Multiple Methods: Solve using two distinct methods.
- (a) $x^2 \frac{dy}{dx} - y^2 = xy$
- (b) $xe^y dy + (e^y + x^3 e^{2x})dx = 0$; $y(1) = 2$
- (c) $(x^2 + 6y^2)dx - 4xydy = 0$; $y(1) = 1$
Combined Methods: The equation $(x^4 + y^4)dx + 4xy^3 dy = 0$ is homogeneous, exact, and Bernoulli. Using all three methods, show the solution is $x^5 + 5xy^4 = A$.

Key Concepts Tested

Identification of Bernoulli differential equations
Transformation: $v = y^{1-n}$ to convert to linear form
Solving linear first-order differential equations
Particular solutions with initial conditions
Homogeneous differential equations
Exact differential equations
Multiple solution methods for verification

Important Formulas

Bernoulli form: $\frac{dy}{dx} + P(x)y = Q(x)y^n$
Substitution: $v = y^{1-n}$, $\frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}$
Linear form after substitution: $\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$
Integrating factor: $\mu(x) = e^{\int P(x)dx}$

st13-detailed