1.3 Bernoulli Differential Equations

Bernoulli differential equations represent an important class of nonlinear equations that can be reduced to linear equations through a clever substitution. Named after Jacob Bernoulli (1654–1705), these equations appear frequently in applications involving population dynamics, fluid mechanics, and chemical reactions.

Learning Objectives

Identify and classify Bernoulli differential equations

Apply the Bernoulli substitution to transform nonlinear equations to linear form

Solve Bernoulli equations and interpret physical solutions

Analyze equilibrium solutions and stability

Standard Form of Bernoulli Equation

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

where: $P(x)$ and $Q(x)$ are continuous functions, $n \neq 0, 1$ is a real constant.

Why Bernoulli equations are important:

Nonlinear yet solvable: While the equation is nonlinear (due to the $y^n$ term), the substitution $v = y^{1-n}$ transforms it into a linear equation.
Extensions of linear theory: They generalize linear equations (when $n=0$) to a broader class of solvable nonlinear ODEs.
Logistic models: When $n=2$, the equation becomes the logistic model: $\frac{dy}{dx} + P(x)y = Q(x)y^2$, fundamental in population biology.

Applications

Population with Harvesting: A population $y(t)$ with growth rate $r$ and carrying capacity $K$ under harvesting rate $h(t)$ follows: $$\frac{dy}{dt} = ry\left(1 - \frac{y}{K}\right) - h(t)$$ For $h(t) = hy^{\alpha}$ (nonlinear harvesting), this becomes a Bernoulli equation.

Fluid Flow (Torricelli's Law): The velocity of fluid draining from a container with nonlinear resistance follows: $$\frac{dv}{dt} + av = bv^3$$ a classic Bernoulli equation in fluid dynamics.

“

Jacob Bernoulli proposed this equation in 1695 and challenged the mathematical community to solve it. His brother Johann, along with Leibniz, found the substitution that bears the family name. The rivalry between the Bernoulli brothers drove some of the greatest advances in early differential equations.

— On the Bernoulli brothers: Jacob (1655–1705) and Johann (1667–1748)

Theory: The Bernoulli Substitution

The key insight to solving Bernoulli equations is to use a substitution that transforms the nonlinear equation into a linear one.

The Transformation Strategy

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

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

Step 2: Differentiate to find $\frac{dv}{dx}$ $$\frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}$$ $$\Rightarrow \frac{dy}{dx} = \frac{1}{1-n}y^n\frac{dv}{dx}$$

Step 3: Substitute into the original equation $$\frac{1}{1-n}y^n\frac{dv}{dx} + P(x)y = Q(x)y^n$$ Divide through by $y^n$ (assuming $y \neq 0$): $$\frac{1}{1-n}\frac{dv}{dx} + P(x)y^{1-n} = Q(x)$$ Since $v = y^{1-n}$: $$\frac{1}{1-n}\frac{dv}{dx} + P(x)v = Q(x)$$

Step 4: Standard linear form $$\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$$

Solving the Linear Equation

The resulting linear equation can be solved using the integrating factor method:

$$\mu(x) = e^{\int (1-n)P(x)\,dx}$$

Multiply through by $\mu(x)$, integrate, and back-substitute $y = v^{1/(1-n)}$.

Important Special Cases

Case 1: $n = 2$ (Logistic-type) $$v = y^{-1}, \quad \frac{dv}{dx} - P(x)v = -Q(x)$$

Case 2: $n = 3$ $$v = y^{-2}, \quad \frac{dv}{dx} + 2P(x)v = 2Q(x)$$

Case 3: $n = -1$ $$v = y^2, \quad \frac{dv}{dx} + 2P(x)v = 2Q(x)$$

Case 4: $n = 1/2$ (Radical case) $$v = y^{1/2}, \quad \frac{dv}{dx} + \frac{1}{2}P(x)v = \frac{1}{2}Q(x)$$

Critical Consideration: The Solution $y = 0$

When we divided by $y^n$ during the substitution, we implicitly assumed $y \neq 0$. Always check whether $y = 0$ is a solution to the original equation:

Substitute $y = 0$ into the original: $\displaystyle 0 + P(x) \cdot 0 = Q(x) \cdot 0$ ✓

Therefore, $y = 0$ is always a solution to any Bernoulli equation. The general solution formula may not capture this singular solution, so report it separately.

Step-by-Step Solution Method (Streamlined Shortcut)

Key Insight: Once you understand the derivation in the Theory section, you can use a direct shortcut. The substitution $v = y^{1-n}$ always transforms a Bernoulli equation into a linear equation with a predictable form. Instead of re-deriving this each time, jump straight to the linear form and solve it.

1 Identify P(x), Q(x), and n

Rewrite the equation in standard form $\displaystyle \frac{dy}{dx} + P(x)y = Q(x)y^n$. Extract the coefficient functions and the exponent.

2 Write the substitution and linear form

Set $v = y^{1-n}$. The key shortcut is that this transformation always yields the linear equation: $$\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$$ (You don't need to re-derive this—it comes directly from the theory. Just substitute your values of $P(x)$, $Q(x)$, and $n$.)

3 Solve the linear DE using integrating factor

Solve the linear equation above using the standard integrating factor formula: $$\mu(x) = e^{\int (1-n)P(x)\,dx}$$ Multiply through by $\mu(x)$ and integrate: $$v = \frac{1}{\mu(x)} \left[ \int \mu(x) \cdot (1-n)Q(x) \, dx + C \right]$$

4 Back-substitute to recover y

Replace $v$ with $y^{1-n}$ and solve for $y$: $$y^{1-n} = v(x, C) \quad \Rightarrow \quad y = v(x, C)^{1/(1-n)}$$ Don't forget: $y = 0$ is always a singular solution (check by substitution).

Memory Aid: Bernoulli $\to$ $v = y^{1-n}$ $\to$ linear form $\to$ integrating factor $\to$ back-substitute. Memorize the pattern of how the linear form is constructed: the coefficient of $v$ becomes $(1-n)P(x)$ and the RHS becomes $(1-n)Q(x)$.

Worked Examples

Example 1: $\displaystyle \frac{dy}{dx} + y = y^2$ (Logistic-type) ▼

Given equation: $\displaystyle \frac{dy}{dx} + y = y^2$

Step 1: Identify parameters. $P(x) = 1$, $Q(x) = 1$, $n = 2$

Step 2: Substitution and linear form. Let $v = y^{1-2} = y^{-1}$. Using the shortcut formula, the linear equation is: $$\frac{dv}{dx} + (1-2)(1) \cdot v = (1-2)(1)$$ $$\frac{dv}{dx} - v = -1$$

Step 3: Solve the linear equation. Integrating factor: $\mu = e^{\int -1 \, dx} = e^{-x}$. $$v = e^x \left[ \int e^{-x}(-1) \, dx + C \right] = e^x \left[ e^{-x} + C \right] = 1 + Ce^x$$

Step 4: Back-substitute. Since $v = y^{-1}$: $$y = \frac{1}{v} = \frac{1}{1 + Ce^x}$$

Solution: $\displaystyle y(x) = \frac{1}{1 + Ce^x}$ (general solution) and $y = 0$ (singular solution).

Example 2: $\displaystyle \frac{dy}{dx} - \frac{2y}{x} = -x^2y^3$ (n = 3) ▼

Given equation: $\displaystyle \frac{dy}{dx} - \frac{2y}{x} = -x^2y^3$

Step 1: Identify parameters. $P(x) = -\frac{2}{x}$, $Q(x) = -x^2$, $n = 3$

Step 2: Substitution and linear form. Let $v = y^{1-3} = y^{-2}$. Using the shortcut formula: $$\frac{dv}{dx} + (1-3)\left(-\frac{2}{x}\right)v = (1-3)(-x^2)$$ $$\frac{dv}{dx} + \frac{4v}{x} = 2x^2$$

Step 3: Solve the linear equation. Integrating factor: $\displaystyle \mu = e^{\int \frac{4}{x}dx} = x^4$. $$v = \frac{1}{x^4} \left[ \int x^4 \cdot 2x^2 \, dx + C \right] = \frac{1}{x^4}\left[\frac{2x^7}{7} + C\right] = \frac{2x^3}{7} + \frac{C}{x^4}$$

Step 4: Back-substitute. Since $v = y^{-2}$: $$y^{-2} = \frac{2x^3}{7} + \frac{C}{x^4} \quad \Rightarrow \quad y = \left(\frac{2x^3}{7} + \frac{C}{x^4}\right)^{-1/2}$$

Solution: $\displaystyle y(x) = \left(\frac{2x^3}{7} + \frac{C}{x^4}\right)^{-1/2}$ and $y = 0$.

Example 3: $\displaystyle xy' + y = x^2y^2, \quad y(1) = 1$ ▼

Given equation: $\displaystyle xy' + y = x^2y^2$ with initial condition $y(1) = 1$.

Step 1: Standard form and identify parameters. Divide by $x$: $$\frac{dy}{dx} + \frac{y}{x} = xy^2$$ So $P(x) = \frac{1}{x}$, $Q(x) = x$, $n = 2$.

Step 2: Substitution and linear form. Let $v = y^{1-2} = y^{-1}$. Using the shortcut formula: $$\frac{dv}{dx} + (1-2)\left(\frac{1}{x}\right)v = (1-2)(x)$$ $$\frac{dv}{dx} - \frac{v}{x} = -x$$

Step 3: Solve the linear equation. Integrating factor: $\mu = e^{\int -\frac{1}{x}dx} = e^{-\ln|x|} = \frac{1}{x}$. $$v = x \left[ \int \frac{1}{x} \cdot (-x) \, dx + C \right] = x \left[ \int -1 \, dx + C \right] = x[-x + C] = -x^2 + Cx$$

Step 4: Apply initial condition. $y(1) = 1 \Rightarrow v(1) = 1$: $$1 = -1 + C \Rightarrow C = 2$$ So $v = -x^2 + 2x = x(2 - x)$.

Step 5: Back-substitute. Since $v = y^{-1}$: $$y = \frac{1}{v} = \frac{1}{x(2-x)}$$

Solution: $\displaystyle y(x) = \frac{1}{x(2-x)}$

Example 4: $\displaystyle \frac{dy}{dx} + \frac{y}{x} = y^{1/2}\ln(x)$ (Fractional power) ▼

Given equation: $\displaystyle \frac{dy}{dx} + \frac{y}{x} = y^{1/2}\ln(x)$

Step 1: Identify parameters. $P(x) = \frac{1}{x}$, $Q(x) = \ln(x)$, $n = \frac{1}{2}$

Step 2: Substitution and linear form. Let $\displaystyle v = y^{1-1/2} = y^{1/2}$. Using the shortcut formula: $$\frac{dv}{dx} + \left(1-\frac{1}{2}\right)\left(\frac{1}{x}\right)v = \left(1-\frac{1}{2}\right)\ln(x)$$ $$\frac{dv}{dx} + \frac{v}{2x} = \frac{\ln(x)}{2}$$

Step 3: Solve the linear equation. Integrating factor: $\mu = e^{\int \frac{1}{2x}dx} = x^{1/2}$. $$v = \frac{1}{\sqrt{x}} \left[ \int \sqrt{x} \cdot \frac{\ln(x)}{2} \, dx + C \right]$$ The integral $\int \sqrt{x}\ln(x)\,dx$ requires integration by parts. The result is implicit:

Step 4: Back-substitute. Since $v = y^{1/2}$, we have $y = v^2$. The implicit solution is: $$\sqrt{x} \, y^{1/2} = \int \sqrt{x} \ln(x) \, dx + C$$

Solution: Given implicitly via the integral of $\sqrt{x}\ln(x)$. This example shows that not all Bernoulli equations have closed-form solutions—sometimes the linear equation itself requires numerical integration.

Example 5: Nonlinear Feedback in Control Systems ▼

Given equation: $\displaystyle \frac{dy}{dt} + 2y = 3y^{1/2}$

In control theory, a system with nonlinear feedback can produce a Bernoulli-type equation. This equation models a system where the output signal $y(t)$ is subject to a restoring force that depends on the square root of the output—common in valve dynamics and hydraulic systems.

Step 1: Identify parameters. $P(t) = 2$, $Q(t) = 3$, $n = \frac{1}{2}$

Step 2: Substitution and linear form. Let $\displaystyle v = y^{1-1/2} = y^{1/2}$. Using the shortcut formula: $$\frac{dv}{dt} + \left(1-\frac{1}{2}\right)(2)v = \left(1-\frac{1}{2}\right)(3)$$ $$\frac{dv}{dt} + v = \frac{3}{2}$$

Step 3: Solve the linear equation. Integrating factor: $\mu = e^{\int 1 \, dt} = e^t$. $$v = e^{-t} \left[ \int e^t \cdot \frac{3}{2} \, dt + C \right] = e^{-t}\left[\frac{3}{2}e^t + C\right] = \frac{3}{2} + Ce^{-t}$$

Step 4: Back-substitute. Since $v = y^{1/2}$, we have $y = v^2$: $$y(t) = \left(\frac{3}{2} + Ce^{-t}\right)^2$$

Solution: $\displaystyle y(t) = \left(\frac{3}{2} + Ce^{-t}\right)^2$ and $y = 0$.

Control Theory Insight: As $t \to \infty$, the transient term $Ce^{-t}$ decays to zero, and the output approaches $y \to \left(\frac{3}{2}\right)^2 = \frac{9}{4}$. This is the steady-state output of the control system—the equilibrium value the system naturally settles to. The Bernoulli substitution transformed a nonlinear control problem into a simple linear ODE, a technique frequently used in nonlinear control theory and systems analysis.

Example 6: $\displaystyle \frac{dy}{dx} + y = e^x y^3$ (Classic n=3 Bernoulli) ▼

Given equation: $\displaystyle \frac{dy}{dx} + y = e^x y^3$

Step 1: Identify parameters. $P(x) = 1$, $Q(x) = e^x$, $n = 3$

Step 2: Substitution and linear form. Let $v = y^{1-3} = y^{-2}$. Using the shortcut formula: $$\frac{dv}{dx} + (1-3)(1)v = (1-3)(e^x)$$ $$\frac{dv}{dx} - 2v = -2e^x$$

Step 3: Solve the linear equation. Integrating factor: $\mu = e^{\int -2 \, dx} = e^{-2x}$. $$v = e^{2x}\left[\int e^{-2x} \cdot (-2e^x) \, dx + C\right] = e^{2x}\left[2e^{-x} + C\right] = 2e^x + Ce^{2x}$$

Step 4: Back-substitute. Since $v = y^{-2}$: $$y^{-2} = 2e^x + Ce^{2x} \quad \Rightarrow \quad y = \pm\frac{1}{\sqrt{2e^x + Ce^{2x}}}$$

Solution: $\displaystyle y(x) = \pm\frac{1}{\sqrt{2e^x + Ce^{2x}}}$ and $y = 0$.

Always remember to substitute back! After solving for $z$, you must return to $y$ using the relationship $z = y^{1-n}$. In this case, since $z = y^{-2}$, we recover $y$ by taking the reciprocal and square root. Forgetting this back-substitution is a common error.

Example 7: $\displaystyle \frac{dy}{dx} - \frac{y}{x} = xy^2$ (n=2 case) ▼

Given equation: $\displaystyle \frac{dy}{dx} - \frac{y}{x} = xy^2$

Step 1: Identify parameters. $P(x) = -\frac{1}{x}$, $Q(x) = x$, $n = 2$

Step 2: Substitution and linear form. Let $v = y^{1-2} = y^{-1}$. Using the shortcut formula: $$\frac{dv}{dx} + (1-2)\left(-\frac{1}{x}\right)v = (1-2)(x)$$ $$\frac{dv}{dx} + \frac{v}{x} = -x$$

Step 3: Solve the linear equation. Integrating factor: $\mu = e^{\int \frac{1}{x}dx} = x$. $$v = \frac{1}{x}\left[\int x \cdot (-x) \, dx + C\right] = \frac{1}{x}\left[-\frac{x^3}{3} + C\right] = -\frac{x^2}{3} + \frac{C}{x}$$

Step 4: Back-substitute. Since $v = y^{-1}$: $$y = \frac{1}{v} = \frac{1}{-\frac{x^2}{3} + \frac{C}{x}} = \frac{3x}{3C - x^3}$$

Solution: $\displaystyle y(x) = \frac{3x}{3C - x^3}$ and $y = 0$.

Why n=2 is special: The case $n=2$ appears frequently in population dynamics (logistic growth, predator-prey models), chemical kinetics (second-order reactions), and fluid mechanics. The transformation to a linear equation is straightforward and the back-substitution is simple.

Example 8: $\displaystyle e^{-x}\frac{dy}{dx} + e^{-x}y = y^3$ (Preprocessing needed) ▼

Given equation: $\displaystyle e^{-x}\frac{dy}{dx} + e^{-x}y = y^3$

Warning: Always normalize first! Notice that the coefficient of $\frac{dy}{dx}$ is not 1. We must divide the entire equation by $e^{-x}$ to put it in standard form before identifying $P(x)$, $Q(x)$, and $n$.

Step 1: Normalize the equation. Divide by $e^{-x}$: $$\frac{dy}{dx} + y = y^3 e^x$$ This reveals standard form with $P(x) = 1$, $Q(x) = e^x$, $n = 3$.

Step 2: Identify parameters and shortcut. This is now identical to Example 6! We have $\frac{dy}{dx} + y = e^x y^3$. Let $v = y^{1-3} = y^{-2}$. Using the shortcut formula: $$\frac{dv}{dx} + (1-3)(1)v = (1-3)(e^x)$$ $$\frac{dv}{dx} - 2v = -2e^x$$

Step 4: Back-substitute. Since $v = y^{-2}$: $$y = \pm\frac{1}{\sqrt{2e^x + Ce^{2x}}}$$

Solution: $\displaystyle y(x) = \pm\frac{1}{\sqrt{2e^x + Ce^{2x}}}$ and $y = 0$.

Preprocessing is Critical: When the coefficient of $\frac{dy}{dx}$ is not 1, you must normalize the equation first by dividing through. This ensures you correctly identify $P(x)$, $Q(x)$, and the nonlinear exponent $n$. Skipping this step leads to incorrect identification of the equation type and wrong solutions. In this example, dividing by $e^{-x}$ is the key insight that reveals the true structure of the problem.

📝 Exam-Style Practice Problems

These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.

Practice 1: $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2 y^3$ (Bernoulli with $n=3$) ▼

Given equation: $\displaystyle \frac{dy}{dx} + \frac{y}{x} = x^2 y^3$

Step 1: Identify parameters. $P(x) = \frac{1}{x}$, $Q(x) = x^2$, $n = 3$

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

Step 3: Substitution. Let $v = y^{1-3} = y^{-2}$. Then $v' = -2y^{-3}y'$, so $y^{-3}y' = -\frac{1}{2}v'$. $$-\frac{1}{2}v' + \frac{v}{x} = x^2$$ $$v' - \frac{2v}{x} = -2x^2$$

Step 4: Solve the linear equation in $v$.

Integrating factor: $\mu = e^{\int -\frac{2}{x}dx} = e^{-2\ln x} = x^{-2}$.

$$\frac{d}{dx}[x^{-2}v] = -2$$

Integrate: $x^{-2}v = -2x + C$

So: $v = -2x^3 + Cx^2$

Step 5: Back-substitute. Since $v = y^{-2}$: $$y^{-2} = -2x^3 + Cx^2$$

Solution: $$\boxed{y^2 = \frac{1}{Cx^2 - 2x^3}} \quad \text{or} \quad \boxed{y = \pm\frac{1}{\sqrt{Cx^2 - 2x^3}}}$$

Singular solution: $y = 0$ also satisfies the equation.

Practice 2: $\dfrac{dy}{dx} - 2y = e^x y^{1/2}$ (Bernoulli with $n = 1/2$) ▼

Given equation: $\displaystyle \frac{dy}{dx} - 2y = e^x y^{1/2}$

Step 1: Identify parameters. $P(x) = -2$, $Q(x) = e^x$, $n = \frac{1}{2}$

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

Step 3: Substitution. Let $v = y^{1-1/2} = y^{1/2}$. Then $v' = \frac{1}{2}y^{-1/2}y'$, so $y^{-1/2}y' = 2v'$. $$2v' - 2v = e^x$$ $$v' - v = \frac{e^x}{2}$$

Step 4: Solve the linear equation in $v$.

Integrating factor: $\mu = e^{\int -1\,dx} = e^{-x}$.

$$\frac{d}{dx}[e^{-x}v] = e^{-x} \cdot \frac{e^x}{2} = \frac{1}{2}$$

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

So: $v = e^x\left(\frac{x}{2} + C\right)$

Step 5: Back-substitute. Since $v = y^{1/2}$: $$\sqrt{y} = e^x\left(\frac{x}{2} + C\right)$$

Solution: $$\boxed{y = \left[e^x\left(\frac{x}{2} + C\right)\right]^2}$$

Singular solution: $y = 0$ (found when $C \to -\infty$ or by direct substitution).

Fractional Exponents: Bernoulli equations with $0 < n < 1$ often produce explicit solutions involving fractional powers and squares. Be careful with domains — ensure the expressions under square roots remain non-negative for your solution to be valid.

Interactive Slope Field Visualizer

Select Example or Custom:

Enter equation (dy/dx = f(x,y)): Use ** for power, * for multiply, / for divide

X range: -1 to 3

Y range: -1 to 3

Click on the plot to trace a solution curve

Parameter Explorer: $\displaystyle \frac{dy}{dx} + ay = by^n$

Parameter a: 1

Parameter b: 1

Exponent n:

Initial condition y₀: 0.5

Behavior Analysis

Adjust the parameters to see how the solutions change.

وَمَا أُوتِيتُمْ مِنَ الْعِلْمِ إِلَّا قَلِيلًا

"And you have not been given of knowledge except a little."

— Quran 17:85

The Bernoulli substitution reminds us that seemingly impossible problems can become simple with the right perspective. In mathematics and in life, a change of viewpoint — a substitution — can transform complexity into clarity. Stay humble before the vastness of knowledge, and keep seeking.

Practice Problems

Self-Assessment Quiz

Question 1: Identify Bernoulli Equations

Which of the following is a Bernoulli equation?

(a) $\displaystyle \frac{dy}{dx} + 2xy = e^x$

(b) $\displaystyle \frac{dy}{dx} + xy = xy^2$

(d) All of the above

Question 2: Identify the Correct Substitution

For the equation $\displaystyle \frac{dy}{dx} + 3y = 2y^4$, what is the correct substitution?

(a) $v = y^{-3}$

(b) $v = y^{-4}$

(d) $v = y^{3}$

Question 3: Transformed Linear Equation

After the substitution in Q2, what is the resulting linear equation in v?

(a) $\displaystyle \frac{dv}{dx} - 9v = -6$

(b) $\displaystyle \frac{dv}{dx} + 9v = 6$

(d) $\displaystyle \frac{dv}{dx} + 3v = 2$

Question 4: General Solution

Solve $\displaystyle \frac{dy}{dx} - y = -y^2$. What is the general solution?

(a) $\displaystyle y = \frac{1}{1 + Ce^x}$

(b) $\displaystyle y = \frac{1}{Ce^x - 1}$

(d) $\displaystyle y = Ce^x$

Question 5: Equilibrium Solutions

What are the equilibrium solutions of $\displaystyle \frac{dy}{dx} + 2y = 5y^3$?

(a) $y = 0, y = \pm\sqrt{2/5}$

(b) $y = 0$ only

(d) $y = 0, y = 2/5$

Question 6: Solve Initial Value Problem

Solve $\displaystyle \frac{dy}{dx} + y = y^2$ with $y(0) = 2$. Find $y(x)$.

(a) $\displaystyle y = \frac{1}{1 - e^x}$

(b) $\displaystyle y = \frac{2}{2 - e^x}$

(d) $\displaystyle y = 2e^x$

Quick Reference Card

Standard Form

$\displaystyle \frac{dy}{dx} + P(x)y = Q(x)y^n$, $n \neq 0, 1$

Bernoulli Substitution

$v = y^{1-n}$

Derivative Relation

$\displaystyle \frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}$

Linear Form

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

Integrating Factor

$\mu(x) = e^{\int (1-n)P(x)\,dx}$

Back-substitution

$y = v^{1/(1-n)}$

Special: n = 2

$v = y^{-1}$, $\displaystyle \frac{dv}{dx} - P(x)v = -Q(x)$

Special: n = 3

$v = y^{-2}$, $\displaystyle \frac{dv}{dx} + 2P(x)v = 2Q(x)$

Always Remember

$y = 0$ is always a solution

Singular Solution

Report $y = 0$ separately from general solution