6.2 Special First-Order ODEs: Homogeneous, Bernoulli, and Orthogonal Trajectories

6.2 Special First-Order ODEs: Homogeneous, Bernoulli, and Orthogonal Trajectories

Many first-order differential equations that are neither separable nor linear can be solved via standard algebraic transformations. The GRE Mathematics Subject Test regularly tests substitution strategies—specifically homogeneous and Bernoulli equations—alongside geometric applications like orthogonal trajectories and theoretical questions on solution uniqueness.

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Homogeneous First-Order Differential Equations

A function $f(x, y)$ is homogeneous of degree $n$ if $f(tx, ty) = t^n f(x, y)$ for all $t > 0$. A first-order ODE $y' = f(x, y)$ is termed homogeneous (or homogeneous of degree zero) if the right-hand side satisfies: $f(tx, ty) = f(x, y) \iff f(x, y) = F\left(\frac{y}{x}\right)$

The Standard Substitution $y = vx$

To solve $y' = F(y/x)$:

1. Define the new dependent variable $v(x) = \frac{y}{x}$, so that $y = v x$.
2. Compute the derivative using the product rule: $\frac{dy}{dx} = v + x\frac{dv}{dx}$
3. Substitute into the differential equation: $v + x\frac{dv}{dx} = F(v) \implies x\frac{dv}{dx} = F(v) - v$
4. This equation is immediately separable: $\frac{dv}{F(v) - v} = \frac{dx}{x}$
5. Integrate both sides and back-substitute $v = y/x$ to obtain the implicit or explicit solution in $x$ and $y$.

Equilibrium Check: Any root $v^$ satisfying $F(v^) - v^* = 0$ corresponds to a straight-line solution $y(x) = v^* x$ passing through the origin.

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Bernoulli Differential Equations

A Bernoulli equation is a classic nonlinear differential equation of the form: $y' + P(x)y = Q(x)y^n, \quad n \in \mathbb{R}$

• If $n = 0$, the equation is standard linear nonhomogeneous: $y' + P(x)y = Q(x)$.
• If $n = 1$, the equation is separable and linear homogeneous: $y' + [P(x) - Q(x)]y = 0$.
• For $n$ not equal to 0 or 1, the equation is genuinely nonlinear.

The Linearizing Substitution $v = y^{1-n}$

1. Divide the entire equation by $y^n$: $y^{-n}y' + P(x)y^{1-n} = Q(x)$
2. Define the substitution variable $v = y^{1-n}$. By the chain rule: $\frac{dv}{dx} = (1 - n)y^{-n}\frac{dy}{dx} \implies y^{-n}y' = \frac{1}{1 - n}\frac{dv}{dx}$
3. Substitute into the differential equation: $\frac{1}{1 - n}v' + P(x)v = Q(x) \implies v' + (1 - n)P(x)v = (1 - n)Q(x)$
4. This is a standard first-order linear ODE in $v(x)$, solved using the integrating factor $\mu(x) = \exp\left((1 - n)\int P(x),dx\right)$.
5. Once $v(x)$ is found, back-substitute $y = v^{1/(1-n)}$. Note that if $n > 0$, $y \equiv 0$ is also a trivial solution.

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Overview of Riccati and Clairaut Equations

Riccati Differential Equations

A Riccati equation is quadratic in the dependent variable: $y' = q_0(x) + q_1(x)y + q_2(x)y^2$ General Riccati equations cannot be solved by elementary quadratures. However, if a particular solution $y_1(x)$ is known or given:

1. Substitute $y(x) = y_1(x) + \frac{1}{u(x)}$.
2. Differentiating and substituting reduces the equation to a first-order linear ODE in $u(x)$: $u' + [q_1(x) + 2 q_2(x) y_1(x)]u = -q_2(x)$

Clairaut Differential Equations

A Clairaut equation has the form: $y = x y' + f(y')$ Differentiating both sides with respect to $x$ gives: $y' = y' + x y'' + f'(y') y'' \implies y'' [x + f'(y')] = 0$

• General Solution: Setting $y'' = 0$ implies $y' = c$ (a constant). Substituting into the original ODE yields the family of straight lines: $y(x) = c x + f(c)$
• Singular Solution: Setting $x + f'(y') = 0$ parameterizes a singular solution (the envelope of the family of lines), which contains no arbitrary constants and cannot be obtained by specializing $c$.

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Orthogonal Trajectories

Given a one-parameter family of planar curves $F(x, y, k) = 0$, its orthogonal trajectories form a second family of curves that intersect every member of the original family at right angles ($90^\circ$).

Systematic Algorithm

1. Find the Governing ODE: Differentiate $F(x, y, k) = 0$ implicitly with respect to $x$.
2. Eliminate Parameter $k$: Solve for $k$ in terms of $x$ and $y$ using $F(x, y, k) = 0$ and substitute it into the differentiated equation, obtaining $y' = f(x, y)$.
3. Replace with Negative Reciprocal: For perpendicular tangents, the orthogonal family's slope must satisfy $m_{\perp} = -\frac{1}{m}$. Replace $y'$ with $-\frac{1}{y'}$ (or equivalently, replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$): $\frac{dy}{dx} = -\frac{1}{f(x, y)}$
4. Integrate: Solve the resulting differential equation to find the orthogonal family $G(x, y, C) = 0$.

Polar Coordinates: In polar coordinates $F(r, \theta, k) = 0$, the tangent angle formula dictates that $\frac{dr}{d\theta}$ is replaced by $-r^2 \frac{d\theta}{dr}$.

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Picard-Lindelöf Existence and Uniqueness Theorem

Consider the initial value problem: $y' = f(x, y), \quad y(x_0) = y_0$

Theorem Statement

Let $R = [x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b]$ be a closed rectangle in $\mathbb{R}^2$ containing $(x_0, y_0)$. If:

1. $f(x, y)$ is continuous on $R$, and
2. $\frac{\partial f}{\partial y}(x, y)$ is continuous on $R$ (which guarantees that $f$ satisfies a uniform Lipschitz condition $|f(x, y_1) - f(x, y_2)| \le L |y_1 - y_2|$ with respect to $y$ on $R$),

then there exists a positive number $h > 0$ such that the IVP has a unique solution $y(x)$ defined on $[x_0 - h, x_0 + h]$.

Breakdown of Uniqueness

• Peano Existence Theorem: Continuity of $f(x, y)$ alone guarantees the existence of at least one local solution, but does not guarantee uniqueness.
• Failure of Lipschitz Continuity: When $\frac{\partial f}{\partial y}$ blows up at $(x_0, y_0)$, uniqueness frequently fails. The prototypical GRE example is $y' = 3 y^{2/3}$ with $y(0) = 0$. Here $\frac{\partial f}{\partial y} = 2 y^{-1/3}$ is undefined at $y = 0$. The IVP possesses infinitely many solutions: $y(x) \equiv 0$ and $y(x) = (x - c)^3$ for $x > c \ge 0$.

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Step-by-Step Worked Problems

Problem 1: Bernoulli Initial Value Problem

Solve the initial value problem $x y' + y = x^2 y^2$ with $y(1) = -1$ on $(0, \infty)$.

Solution:

1. Express in standard Bernoulli form ($n = 2$): $y' + \frac{1}{x}y = x y^2$
2. Divide by $y^2$: $y^{-2}y' + \frac{1}{x}y^{-1} = x$
3. Set $v = y^{1-2} = y^{-1}$, giving $v' = -y^{-2}y'$: $-v' + \frac{1}{x}v = x \implies v' - \frac{1}{x}v = -x$
4. Find the integrating factor: $\mu(x) = \exp\left(\int -\frac{1}{x},dx\right) = \frac{1}{x}$.
5. Multiply and integrate: $\frac{d}{dx}\left[\frac{v}{x}\right] = \frac{-x}{x} = -1 \implies \frac{v}{x} = -x + C \implies v(x) = -x^2 + Cx$
6. Invert $v = 1/y$ to find $y(x)$: $y(x) = \frac{1}{Cx - x^2}$
7. Apply initial condition $y(1) = -1$: $\frac{1}{C - 1} = -1 \implies C - 1 = -1 \implies C = 0$
8. The explicit solution is $y(x) = -\frac{1}{x^2}$.

Problem 2: Orthogonal Trajectories of Parabolas

Find the orthogonal trajectories to the family of parabolas $y = c x^2$.

Solution:

1. Differentiate implicitly: $y' = 2 c x$.
2. Eliminate $c$: From $y = c x^2$, we have $c = \frac{y}{x^2}$. Substituting gives: $y' = 2 \left(\frac{y}{x^2}\right) x = \frac{2y}{x}$
3. Replace $y'$ with $-\frac{1}{y'}$ for the orthogonal family: $-\frac{1}{y'} = \frac{2y}{x} \implies \frac{dy}{dx} = -\frac{x}{2y}$
4. Separate variables and integrate: $2y\,dy = -x\,dx \implies y^2 = -\frac{1}{2}x^2 + C_0 \implies \frac{x^2}{2} + y^2 = C_0$
5. This defines a family of concentric ellipses centered at the origin: $x^2 + 2y^2 = C$.

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GRE Exam Traps & Pitfalls

Trap 1: Forgetting Negative Reciprocal in Orthogonal Trajectories Replacing $y'$ with $-y'$ merely reflects the slope across the horizontal; orthogonal trajectories require the negative reciprocal $-\frac{1}{y'}$. In polar coordinates, replacing $r'$ with $-1/r'$ is incorrect; the polar relation requires replacing $\frac{dr}{d\theta}$ with $-r^2 \frac{d\theta}{dr}$.

Trap 2: Sign Errors in Bernoulli Substitution When substituting $v = y^{1-n}$, remember that $v' = (1-n)y^{-n}y'$. Missing the factor of $(1-n)$ leads to incorrect integrating factors and solutions.

Trap 3: Confusing Existence with Uniqueness Peano's Theorem guarantees that continuity of $f(x, y)$ alone is sufficient for a solution to exist. Picard-Lindelöf requires continuity of $\frac{\partial f}{\partial y}$ for uniqueness. Questions on the GRE test whether you can identify that non-Lipschitz behavior permits multiple solutions while maintaining existence.