Bernoulli Equations - Differential Equations

Consider the modified Bernoulli equation $\frac{dy}{dx} + P(x)y = Q(x)y^n + R(x)y^m$ where $n \neq 1$ , $m \neq 1$ , and $n \neq m$ . Under what condition can this equation still be solved using a single Bernoulli-type substitution?
Answer: When $R(x) = 0$ (reducing to standard Bernoulli) or when $Q(x) = 0$ When you encounter a modified Bernoulli equation with two nonlinear terms, the key insight is recognizing when it can be reduced to a solvable form using standard techniques. The correct answer is A because a single Bernoulli-type substitution only works when the equation effectively reduces to the standard Bernoulli form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ . This happens when either $R(x) = 0$ (eliminating the $y^m$ term entirely) or $Q(x) = 0$ (eliminating the $y^n$ term). In both cases, you're left with a standard Bernoulli equation that can be solved using the substitution $v = y^{1-n}$ or $v = y^{1-m}$ respectively. Option B is incorrect because the condition $m = 2n - 1$ doesn't eliminate the fundamental problem of having two different nonlinear terms. The substitution $v = y^{1-n}$ would still leave you with terms involving different powers of $v$ . Option C fails because even when $n + m = 1$ , the proposed substitution $v = y^{(1-n)/2}$ doesn't transform both nonlinear terms into the same power simultaneously. You'd still have a mixed nonlinear equation. Option D is wrong because if $n = 0$ or $m = 0$ , those terms become linear ( $y^0 = 1$ ), but you still have one remaining nonlinear term plus the linear $y$ term, which isn't a standard Bernoulli equation. Study tip: Remember that Bernoulli equations can only handle one nonlinear term at a time. When you see multiple nonlinear terms, look for conditions that eliminate all but one of them.
What is the general solution to the differential equation $y' - y = e^x y^2$ ?
Answer: $y(x) = \frac{2e^x}{C - e^{2x}}$ The equation is a Bernoulli equation with $P(x)=-1$ , $Q(x)=e^x$ , and $n=2$ . The substitution is $v = y^{1-2} = y^{-1}$ . Then $v' = -y^{-2}y'$ . Multiplying the original equation by $-y^{-2}$ gives $-y^{-2}y' + y^{-1} = -e^x$ . This transforms into the linear equation $v' + v = -e^x$ . The integrating factor is $\mu(x) = e^{\int 1 dx} = e^x$ . Multiplying the linear equation by $\mu(x)$ gives $(e^x v)' = -e^x \cdot e^x = -e^{2x}$ . Integrating both sides yields $e^x v = \int -e^{2x} dx = -\frac{1}{2}e^{2x} + C_1$ . Solving for $v$ gives $v = -\frac{1}{2}e^x + C_1e^{-x}$ . Substituting back $v = y^{-1}$ : $y^{-1} = \frac{-e^{2x} + 2C_1}{2e^x}$ . Inverting to find $y$ gives $y(x) = \frac{2e^x}{2C_1 - e^{2x}}$ . Renaming the constant $C=2C_1$ , we get $y(x) = \frac{2e^x}{C - e^{2x}}$ . Distractor B results from a sign error when integrating $-e^{2x}$ .
For the Bernoulli equation $\frac{dy}{dx} - \frac{2y}{x} = \frac{y^4}{x^3}$ , suppose we make an error and use the substitution $v = y^{-4}$ instead of the correct Bernoulli substitution. What type of equation would we obtain for $v$ ?
Answer: Another Bernoulli equation with a different power The correct Bernoulli substitution should be v = $y^{1-4}$ = $y^{-3}$ . If we incorrectly use v = $y^{-4}$ , then y = $v^{-1/4}$ and dy/dx = (-1/4) $v^{-5/4}$ dv/dx. Substituting into the original equation: (-1/4) $v^{-5/4}$ dv/dx - $2v^{-1/4}$ /x = $v^{-1}$ / $x^3$ . Multiplying by - $4v^{5/4}$ : dv/dx + 8v/x = - $4v^{1/4}$ / $x^3$ . This is still a Bernoulli equation with power 1/4, not linear. Choice A would be correct for the proper substitution. Choice B is wrong since this maintains the differential equation structure. Choice D is incorrect as the order doesn't change.
Consider the differential equation $y' \cos(x) + y \sin(x) = y^3 \tan(x)$ . Given the initial condition $y(0)=1$ , which expression defines the solution $y(x)^2$ for $x$ in the domain of the solution?
Answer: $\frac{\cos^2(x)}{2\cos(x) - 1}$ When you encounter a differential equation where the coefficients suggest a product rule pattern, look for opportunities to use substitution techniques. This equation has the form that suggests rewriting it to reveal a hidden structure. First, divide the entire equation by $\cos(x)$ : $y' + y\tan(x) = y^3 \sec(x)\tan(x)$ Notice that the left side is the derivative of $y\sec(x)$ by the product rule. Let $v = y\sec(x)$ , so $v' = y'\sec(x) + y\sec(x)\tan(x)$ . Multiplying our equation by $\sec(x)$ : $v' = y^3\sec^2(x)\tan(x)$ Since $y = v\cos(x)$ , we have $y^3 = v^3\cos^3(x)$ , giving us: $v' = v^3\cos(x)\tan(x) = v^3\sin(x)$ This is a separable Bernoulli equation. Using the substitution $w = v^{-2}$ , we get $w' = -2v^{-3}v' = -2\sin(x)$ . Integrating: $w = 2\cos(x) + C$ . With $y(0) = 1$ , we have $v(0) = 1$ , so $w(0) = 1 = 2 + C$ , giving $C = -1$ . Therefore $w = 2\cos(x) - 1$ , and since $w = v^{-2} = (y\sec(x))^{-2} = y^{-2}\cos^2(x)$ : $y^2 = \frac{\cos^2(x)}{2\cos(x) - 1}$ This confirms answer C is correct. Option A has the wrong functional form entirely. Option B has $2\cos(x) + 1$ instead of the correct $2\cos(x) - 1$ , likely from a sign error in the integration constant. Option D has $3 - 2\cos(x)$ , suggesting an error in the Bernoulli substitution. Strategy tip: When you see trigonometric coefficients that look like derivatives of products, try substitutions that exploit the product rule structure before attempting direct integration.
The Bernoulli equation $3y' - xy = \frac{x}{y^2}$ can be transformed into a first-order linear differential equation in the variable $v$ using the substitution $v = y^3$ . What is the resulting linear equation?
Answer: $\frac{dv}{dx} - xv = x$ First, write the given equation in standard Bernoulli form, $y' + P(x)y = Q(x)y^n$ . Divide the entire equation by 3: $y' - \frac{x}{3}y = \frac{x}{3}y^{-2}$ . This is a Bernoulli equation with $P(x) = -x/3$ , $Q(x) = x/3$ , and $n=-2$ . The substitution is $v = y^{1-n} = y^{1-(-2)} = y^3$ . Differentiate $v$ with respect to $x$ : $\frac{dv}{dx} = 3y^2 \frac{dy}{dx}$ . Now, multiply the original equation, $3y' - xy = \frac{x}{y^2}$ , by $y^2$ to get $3y^2y' - xy^3 = x$ . Substitute $v = y^3$ and $v' = 3y^2y'$ into this modified equation. This yields $\frac{dv}{dx} - xv = x$ . This is the correct transformed linear equation.
The population $P(t)$ of a species is initially governed by the logistic equation $\frac{dP}{dt} = 0.2P - 0.004P^2$ . A new disease is introduced, which causes the rate of change of the population to decrease by an amount proportional to the square of the population, with a proportionality constant of $0.001$ . What is the new long-term stable population for this species, assuming $P(0) > 0$ ?
Answer: 40 The original rate of change is $\frac{dP}{dt} = 0.2P - 0.004P^2$ . The new disease adds a death term, decreasing the rate by $0.001P^2$ . The new differential equation is $\frac{dP_{new}}{dt} = (0.2P - 0.004P^2) - 0.001P^2 = 0.2P - 0.005P^2$ . This is a Bernoulli equation (and also a logistic equation). The long-term stable population is the non-zero equilibrium point. We find the equilibria by setting $\frac{dP}{dt} = 0$ : $0.2P - 0.005P^2 = 0 \implies P(0.2 - 0.005P) = 0$ . The equilibrium points are $P=0$ and $P = \frac{0.2}{0.005} = \frac{200}{5} = 40$ . Since the initial population $P(0) > 0$ , the population will approach the stable non-zero equilibrium, which is $P=40$ . Distractor A (50) is the original stable population before the disease. Distractor C (200) results from a decimal error. Distractor D (20) would be the result if the new disease term was $0.006P^2$ instead of $0.001P^2$ , leading to a total quadratic term of $-0.01P^2$ .
Consider the Bernoulli differential equation $x^2 y' + 2xy = y^3$ . A substitution is made to transform this into a linear differential equation. What is the integrating factor, $\mu(x)$ , for the resulting linear equation?
Answer: $x^{-4}$ First, write the equation in standard Bernoulli form: $y' + \frac{2}{x}y = \frac{1}{x^2}y^3$ . Here, $P(x) = \frac{2}{x}$ and $n=3$ . The substitution is $v = y^{1-n} = y^{1-3} = y^{-2}$ . The resulting linear equation has the form $v' + (1-n)P(x)v = (1-n)Q(x)$ . Substituting the values, we get: $v' + (1-3)\frac{2}{x}v = (1-3)\frac{1}{x^2}$ , which simplifies to $v' - \frac{4}{x}v = -\frac{2}{x^2}$ . This is a linear first-order equation in $v$ . The integrating factor is $\mu(x) = e^{\int P_v(x)dx}$ , where $P_v(x) = -\frac{4}{x}$ . So, $\mu(x) = e^{\int -\frac{4}{x} dx} = e^{-4\ln|x|} = e^{\ln(x^{-4})} = x^{-4}$ . Distractor A, $x^2$ , is the integrating factor for the original equation if it were linear. Distractor B, $x^4$ , results from a sign error in the exponent of the integrating factor. Distractor D, $x^{-6}$ , comes from incorrectly calculating the new coefficient as $n P(x) = 3(2/x)$ instead of $(1-n)P(x)$ .
For the Bernoulli equation $\frac{dy}{dt} + ay = by^n$ where $a, b$ are positive constants and $n \neq 0, 1$ , which substitution transforms it into a linear differential equation, and what condition must be satisfied for the transformed equation to have a decreasing solution when the original equation has $y > 0$ ?
Answer: $v = y^{1-n}$ ; requires $a(1-n) > 0$ and $b(1-n) < 0$ The standard Bernoulli substitution is v = $y^{1-n}$ , which transforms the equation to dv/dt + a(1-n)v = b(1-n). For a decreasing solution (dv/dt < 0), we need the coefficient of v to be positive and the right-hand side to be negative when v > 0. This requires a(1-n) > 0 and b(1-n) < 0. Choice A has wrong sign for the b(1-n) condition. Choice C uses wrong substitution v = $y^{n-1}$ . Choice D doesn't properly account for the sign change needed in the b term.
Find the value of $y(1)$ for the solution to the initial value problem $y' = y(xy^3 - 1)$ with the initial condition $y(0) = \frac{1}{2}$ .
Answer: $\left(\frac{3}{4+23e^3}\right)^{1/3}$ First, rewrite the differential equation in the standard Bernoulli form: $y' + y = xy^4$ . Here, $P(x)=1$ , $Q(x)=x$ , and $n=4$ . The substitution is $v = y^{1-4} = y^{-3}$ . This implies $v' = -3y^{-4}y'$ . Multiply the standard form equation by $-3y^{-4}$ : $-3y^{-4}y' - 3y^{-3} = -3x$ . This becomes the linear equation $v' - 3v = -3x$ . The integrating factor is $\mu(x) = e^{\int -3 dx} = e^{-3x}$ . Multiplying the linear equation by $\mu(x)$ gives $(e^{-3x}v)' = -3xe^{-3x}$ . Integrate the right side using integration by parts: $\int -3xe^{-3x} dx = xe^{-3x} + \frac{1}{3}e^{-3x} + C$ . So, $e^{-3x}v = xe^{-3x} + \frac{1}{3}e^{-3x} + C$ , which means $v(x) = x + \frac{1}{3} + Ce^{3x}$ . Substitute back $v=y^{-3}$ : $y^{-3} = x + \frac{1}{3} + Ce^{3x}$ . Now use the initial condition $y(0) = 1/2$ : $(1/2)^{-3} = 8 = 0 + \frac{1}{3} + C$ , so $C = 8 - \frac{1}{3} = \frac{23}{3}$ . The particular solution is $y^{-3} = x + \frac{1}{3} + \frac{23}{3}e^{3x}$ . Finally, evaluate at $x=1$ : $y(1)^{-3} = 1 + \frac{1}{3} + \frac{23}{3}e^3 = \frac{4}{3} + \frac{23}{3}e^3 = \frac{4+23e^3}{3}$ . Therefore, $y(1) = \left(\frac{3}{4+23e^3}\right)^{1/3}$ .
Consider the family of Bernoulli equations $\frac{dy}{dx} + \frac{y}{x} = Cy^n$ where $C$ is a positive constant and $n > 1$ . For which values of $n$ will the substituted linear equation have solutions that can be expressed in terms of elementary functions without requiring special functions or infinite series?
Answer: For any rational $n > 1$ where $1-n$ has denominator 1 or 2 The Bernoulli substitution v = $y^{1-n}$ transforms the equation to dv/dx + (1-n)v/x = C(1-n). This has integrating factor μ = $x^{1-n}$ , giving solution v = Cx + $Dx^{n-1}$ . To express y = $v^{1/(1-n)}$ in elementary functions, we need 1/(1-n) to be a rational number that doesn't require roots with denominators larger than 2 (since higher roots aren't elementary). This happens when 1-n has denominator 1 or 2. Choice A is too restrictive about odd denominators. Choice B only considers integer n. Choice D is incorrect since some rational exponents lead to non-elementary functions.
Consider the equation $y' + \frac{y}{x} = \frac{y^3}{x^3}$ for $x > 0$ . After making the appropriate Bernoulli substitution and solving, if we require that $\lim_{x \to \infty} y(x) = 0$ , what constraint does this place on the arbitrary constant?
Answer: The constant can be any positive value and the solution approaches zero like $x^{-1}$ This is Bernoulli with n = 3. Using v = $y^{-2}$ , the equation becomes dv/dx - (2/x)v = -2/ $x^3$ . The integrating factor is μ = $x^{-2}$ , giving d/dx( $x^{-2}$ v) = - $2x^{-5}$ . Integrating: $x^{-2}$ v = (1/2) $x^{-4}$ + C, so v = (1/2) $x^{-2}$ + $Cx^2$ . Thus $y^{-2}$ = (1/2) $x^{-2}$ + $Cx^2$ , giving y = 1/√((1/2) $x^{-2}$ + $Cx^2$ ). For y → 0 as x → ∞, we need the denominator to grow, which happens when C > 0 (the $Cx^2$ term dominates). The solution behaves like y ≈ 1/(Cx) = (1/C) $x^{-1}$ for large x. Choice A has wrong decay rate. Choice B requires C < 0 which gives unbounded solutions. Choice C eliminates the interesting non-trivial solutions.
The substitution $v=y^{-3}$ transforms a certain Bernoulli equation $\frac{dy}{dx} + P(x)y = Q(x)y^n$ into the linear equation $\frac{dv}{dx} - \frac{9}{x}v = -3x^2$ . What is the original Bernoulli equation?
Answer: $\frac{dy}{dx} + \frac{3}{x}y = x^2 y^4$ The substitution is $v=y^{-3}$ . In the general theory of Bernoulli equations, the substitution is $v=y^{1-n}$ . Thus, $1-n = -3$ , which implies $n=4$ . The transformed linear equation is $v' + (1-n)P(x)v = (1-n)Q(x)$ . Substituting $1-n=-3$ , we get $v' - 3P(x)v = -3Q(x)$ . We are given that the transformed equation is $\frac{dv}{dx} - \frac{9}{x}v = -3x^2$ . By comparing the coefficients of $v$ and the right-hand side terms, we can set up a system of equations: $-3P(x) = -\frac{9}{x}$ and $-3Q(x) = -3x^2$ . Solving for $P(x)$ and $Q(x)$ gives $P(x) = \frac{3}{x}$ and $Q(x) = x^2$ . The original Bernoulli equation is $\frac{dy}{dx} + P(x)y = Q(x)y^n$ , which is $\frac{dy}{dx} + \frac{3}{x}y = x^2 y^4$ .
Determine the largest interval of existence for the solution to the initial value problem $y' - \frac{3}{2x}y = -\frac{1}{2}x y^3$ , with $y(1)=1$ .
Answer: $(0, \infty)$ When you encounter an initial value problem with a nonlinear differential equation like this one, you need to determine where the solution can exist without encountering singularities or breakdown points. This is a Bernoulli equation of the form $y' - \frac{3}{2x}y = -\frac{1}{2}x y^3$ . The key insight is that the coefficient $\frac{3}{2x}$ has a singularity at $x = 0$ , which immediately restricts our domain. Since we have the initial condition $y(1) = 1$ , we're starting at a positive $x$ -value. For Bernoulli equations, we can transform this into a linear equation by substituting $v = y^{-2}$ . After this transformation and solving, we get a solution that exists as long as the denominator doesn't become zero and we stay away from the singularity at $x = 0$ . The solution can be extended throughout the entire positive real line without encountering any breakdown points. Choice A, $(-\sqrt[5]{4}, \infty)$ , incorrectly suggests the solution can exist for some negative $x$ -values, but the singularity at $x = 0$ prevents this. Choice B, $(-\infty, \infty)$ , makes the same error by ignoring the singularity entirely. Choice C, $(1, \infty)$ , is too restrictive—while the initial condition is at $x = 1$ , the solution can actually be extended to all positive values, including those between 0 and 1. The correct answer is D: $(0, \infty)$ . Study tip: For differential equations with coefficients that have singularities, always check where those singularities occur—they typically form boundaries for your interval of existence.
Given that $y_1(x) = \frac{1}{x}$ is a particular solution to the Riccati equation $y' = \frac{1}{x^2} - \frac{y}{x} - y^2$ , the substitution $y = y_1 + u$ transforms it into a Bernoulli equation for $u(x)$ . What is the resulting general solution for $y(x)$ ?
Answer: $y(x) = \frac{C x^2 + 1}{x(C x^2 - 1)}$ When you encounter a Riccati equation with a known particular solution, the standard approach is to use the substitution $y = y_1 + u$ to transform it into a solvable Bernoulli equation. Starting with the given Riccati equation $y' = \frac{1}{x^2} - \frac{y}{x} - y^2$ and the particular solution $y_1(x) = \frac{1}{x}$ , substitute $y = \frac{1}{x} + u$ . Since $y' = -\frac{1}{x^2} + u'$ , the equation becomes: $-\frac{1}{x^2} + u' = \frac{1}{x^2} - \frac{\frac{1}{x} + u}{x} - \left(\frac{1}{x} + u\right)^2$ Expanding and simplifying (noting that $y_1$ satisfies the original equation), you get: $u' = -\frac{u}{x} - \frac{2u}{x} - u^2 = -\frac{3u}{x} - u^2$ This is a Bernoulli equation: $u' + \frac{3u}{x} = -u^2$ . Using the standard Bernoulli substitution $v = u^{-1}$ , this transforms to a linear equation whose solution yields $u = \frac{x}{Cx^3 - 1}$ . Therefore: $y = \frac{1}{x} + \frac{x}{Cx^3 - 1} = \frac{Cx^3 - 1 + x^2}{x(Cx^3 - 1)} = \frac{Cx^2 + 1}{x(Cx^2 - 1)}$ Answer B is correct. Answer A uses the wrong sign and coefficient in the denominator. Answer C has the numerator and denominator terms reversed. Answer D has an incorrect sign in the particular solution addition. Remember: Riccati-to-Bernoulli transformations always follow this systematic substitution pattern—master the algebraic steps to avoid sign errors that create these common distractors.
The Bernoulli equation $y' + \alpha x^2 y = \beta x^5 y^4$ is transformed by a substitution $v=y^k$ into the linear equation $v' - 6x^2 v = -9x^5$ . Determine the values of the constants $\alpha$ , $\beta$ , and $k$ .
Answer: $\alpha=2, \beta=3, k=-3$ When you encounter a Bernoulli equation of the form $y' + P(x)y = Q(x)y^n$ , the standard approach is to use the substitution $v = y^{1-n}$ to transform it into a linear equation. Understanding this transformation process is key to solving these problems. Starting with $y' + \alpha x^2 y = \beta x^5 y^4$ , we need to determine what substitution $v = y^k$ produces the given linear equation $v' - 6x^2 v = -9x^5$ . If $v = y^k$ , then $v' = ky^{k-1}y'$ , so $y' = \frac{v'}{ky^{k-1}}$ . Since $y^k = v$ , we have $y = v^{1/k}$ , giving us $y' = \frac{v'}{kv^{(k-1)/k}}$ . Substituting into the original equation and simplifying, we need the coefficient of $v$ to be $-6x^2$ and the right side to be $-9x^5$ . Working through the algebra with $k = -3$ : - The $y$ term becomes $\alpha x^2 v^{-1/3} \cdot v^{1/3} = \alpha x^2$ - For this to equal $-6x^2$ , we need $\alpha = 2$ (accounting for sign changes) - Similarly, analyzing the $y^4$ term gives us $\beta = 3$ Option A gives the wrong signs for $\alpha$ and $\beta$ . Option B has the correct $k$ but incorrect magnitudes for $\alpha$ and $\beta$ . Option C has the wrong sign for $k$ , which would lead to an incorrect transformation. Study tip: Always verify Bernoulli substitutions by working backwards—substitute your proposed values into the transformation and check that you recover the target linear equation.

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