Differential Equations Quiz: Bernoulli Equations

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Bernoulli EquationsQuestion 1 of 15

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 $y(1)=1$ .

(-\sqrt[5]{4}, \infty)

(-\infty, \infty)

(1, \infty)

(0, \infty)

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Differential Equations Quiz

Differential Equations Quiz: Bernoulli Equations

Practice Bernoulli Equations in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Bernoulli Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

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 $y(1)=1$ .

$(-\sqrt[5]{4}, \infty)$
$(-\infty, \infty)$
$(1, \infty)$
$(0, \infty)$ (correct answer)

Explanation: 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.

Question 2

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 $y > 0$ ?

$v = y^{1-n}$ ; requires $a(1-n) > 0$ and $b(1-n) > 0$
$v = y^{1-n}$ ; requires $a(1-n) > 0$ and $b(1-n) < 0$ (correct answer)
$v = y^{n-1}$ ; requires $a(n-1) > 0$ and $b(n-1) > 0$
$v = y^{1-n}$ ; requires $a > 0$ and $b(1-n) > 0$

Explanation: 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.

Question 3

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?

$\frac{dv}{dx} - xv = x$ (correct answer)
$\frac{dv}{dx} - \frac{x}{3}v = \frac{x}{3}$
$\frac{dv}{dx} + xv = x$
$\frac{dv}{dx} - 3xv = 3x$

Explanation: 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.

Question 4

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 $P(0) > 0$ ?

50
40 (correct answer)
200
20

Explanation: 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

Question 5

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?

$\frac{dy}{dx} + \frac{3}{x}y = x^2 y^4$ (correct answer)
$\frac{dy}{dx} - \frac{3}{x}y = -x^2 y^4$
$\frac{dy}{dx} + \frac{9}{4x}y = \frac{3x^2}{4} y^{-3}$
$\frac{dy}{dx} + 3xy = x^2 y^4$

Explanation: 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

Question 6

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?

$x^2$
$x^4$
$x^{-4}$ (correct answer)
$x^{-6}$

Explanation: 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)

Question 7

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 $y(x)$ ?

$y(x) = \frac{1}{x} + \frac{1}{Cx^3 - x/2}$
$y(x) = \frac{C x^2 + 1}{x(C x^2 - 1)}$ (correct answer)
$y(x) = \frac{C x^2 - 1}{x(C x^2 + 1)}$
$y(x) = \frac{1}{x} + \frac{1}{Cx^3 + x/2}$

Explanation: 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.

Question 8

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 $k$ .

$\alpha=-6, \beta=-9, k=-3$
$\alpha=-2, \beta=-3, k=-3$
$\alpha=2, \beta=3, k=3$
$\alpha=2, \beta=3, k=-3$ (correct answer)

Explanation: 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.

Question 9

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?

The constant must be positive and the solution approaches zero exponentially
The constant must be negative and the solution approaches zero like $x^{-1}$
The constant must be zero and the solution is identically the trivial solution
The constant can be any positive value and the solution approaches zero like $x^{-1}$ (correct answer)

Explanation: 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.

Question 10

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?

Only when $n$ is a rational number with odd denominator
Only when $n = 2, 3, 4, 5, ...$ (positive integers greater than 1)
For any rational $n > 1$ where $1-n$ has denominator 1 or 2 (correct answer)
For all rational $n > 1$ since the Bernoulli substitution always yields elementary solutions

Explanation: 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.

Question 11

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?

When $R(x) = 0$ (reducing to standard Bernoulli) or when $Q(x) = 0$ (correct answer)
When $m = 2n - 1$ and we use substitution $v = y^{1-n}$
When $n + m = 1$ and we use substitution $v = y^{(1-n)/2}$
When either $n = 0$ or $m = 0$ , making one term linear

Explanation: 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}

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

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.

Question 12

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 $v$ ?

A linear first-order equation with constant coefficients
A separable equation that is not linear
Another Bernoulli equation with a different power (correct answer)
A second-order linear equation due to the transformation error

Explanation: 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.

Question 13

What is the general solution to the differential equation $y' - y = e^x y^2$ ?

$y(x) = \frac{2e^x}{C - e^{2x}}$ (correct answer)
$y(x) = \frac{2e^x}{C + e^{2x}}$
$y(x) = \frac{e^{-x}}{C - \frac{1}{2}e^{x}}$
$y(x) = \frac{1}{x e^x + C e^x}$

Explanation: 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}

Question 14

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?

$\frac{1}{\sec(x) - \tan(x)}$
$\frac{\cos^2(x)}{2\cos(x) + 1}$
$\frac{\cos^2(x)}{2\cos(x) - 1}$ (correct answer)
$\frac{\cos^2(x)}{3-2\cos(x)}$

Explanation: 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.

Question 15

Find the value of $y(1)$ for the solution to the initial value problem $y' = y(xy^3 - 1)$ with the initial condition $y(0) = \frac{1}{2}$ .

$\left(\frac{3}{4+23e^3}\right)^{1/3}$ (correct answer)
$\left(\frac{3}{2+23e^3}\right)^{1/3}$
$\left(\frac{3}{4-23e^3}\right)^{1/3}$
$-\left(\frac{21}{4e^3}\right)^{1/3}$

Explanation: 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}