Differential Equations Quiz: Ivps And Existence Uniqueness

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Ivps And Existence UniquenessQuestion 1 of 10

Consider the initial value problem $y' = |y|^{1/3}$ , $y(x_0) = y_0$ . The Existence and Uniqueness Theorem guarantees a unique solution passing through $(x_0, y_0)$ provided that the function $f(x,y)=|y|^{1/3}$ and its partial derivative $\frac{\partial f}{\partial y}$ are continuous in a rectangle around $(x_0, y_0)$ . For which of the following initial conditions is a unique solution not guaranteed by the theorem?

y(1) = 1

y(1) = -1

y(0) = 0

y(-1) = 8

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Differential Equations Quiz

Differential Equations Quiz: Ivps And Existence Uniqueness

Practice Ivps And Existence Uniqueness 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 Ivps And Existence Uniqueness, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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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

$y(1) = 1$
$y(1) = -1$
$y(0) = 0$ (correct answer)
$y(-1) = 8$

Explanation: The function

f(x,y) = |y|^{1/3}

is continuous for all

(x,y)

, so existence of a solution is guaranteed everywhere. For uniqueness, we examine the partial derivative with respect to

y

. For

y > 0

\frac{\partial f}{\partial y} = \frac{1}{3}y^{-2/3}

. For

y < 0

f(x,y) = (-y)^{1/3}

, so

\frac{\partial f}{\partial y} = \frac{1}{3}(-y)^{-2/3}(-1) = -\frac{1}{3}(-y)^{-2/3}

. As

y \to 0

from either the positive or negative side, the magnitude of

\frac{\partial f}{\partial y}

approaches infinity. Thus,

\frac{\partial f}{\partial y}

is not continuous at any point where

y=0

. The initial condition

y(0)=0

has

y_0=0

, so the theorem does not guarantee uniqueness. For all other choices,

y_0 \neq 0

, and

\frac{\partial f}{\partial y}

is continuous in a neighborhood of the initial point.

Question 2

Consider the IVP $y' = \frac{\ln(x+5)}{y-3}$ , $y(0)=1$ . The Existence and Uniqueness Theorem guarantees a unique solution exists in some neighborhood of $x=0$ . The conditions of the theorem are satisfied on any open rectangle containing the initial point $(0,1)$ where $f(x,y)$ and $\frac{\partial f}{\partial y}$ are continuous. What is the largest open interval of $x$ -values, containing $x_0=0$ , that is part of such a region of continuity?

$(-5, \infty)$ (correct answer)
$(-5, 3)$
$(0, 3)$
$(-h, h)$ for some small $h>0$ , the exact value of which requires solving the equation.

Explanation: Let

f(x,y) = \frac{\ln(x+5)}{y-3}

. For

f(x,y)

to be continuous, we need the argument of the logarithm to be positive,

x+5 > 0 \implies x > -5

, and the denominator to be non-zero,

y \neq 3

. The partial derivative is

\frac{\partial f}{\partial y} = -\frac{\ln(x+5)}{(y-3)^2}

, which has the same continuity requirements. The initial point is

(0,1)

. The region of continuity containing

(0,1)

is defined by

x > -5

and

y < 3

. The question asks for the largest open interval for

x

containing

x_0=0

where the conditions hold. This interval is determined solely by the constraint on

x

, which is

x > -5

. Therefore, the largest such interval is

(-5, \infty)

. Distractor D confuses the region where the theorem's conditions hold with the guaranteed interval of existence for the solution, which is only local.

Question 3

Consider the initial value problem $y' = f(x,y)$ , $y(0)=1$ , where the function $f(x,y)$ is defined piecewise:

Which statement accurately describes the existence and uniqueness of a solution as guaranteed by the Picard-Lindelöf theorem?

A solution is guaranteed to exist, but it is not guaranteed to be unique. (correct answer)
A unique solution is guaranteed to exist in a neighborhood of $x=0$ .
No solution is guaranteed to exist because $f(x,y)$ is discontinuous at $y=1$ .
A solution is guaranteed to exist, but only for $x > 0$ .

Explanation: For existence, we check the continuity of

f(x,y)

at the initial point

(0,1)

. As

y \to 1^+

f(x,y) \to x^2

. As

y \to 1^-

f(x,y) \to (1)-1+x^2 = x^2

. Since the limits are equal,

f(x,y)

is continuous everywhere. Therefore, the theorem guarantees a solution exists. For uniqueness, we check the continuity of

\frac{\partial f}{\partial y}

. For

y > 1

\frac{\partial f}{\partial y} = 0

. For

y < 1

\frac{\partial f}{\partial y} = 1

. Since

\lim_{y \to 1^+} \frac{\partial f}{\partial y} = 0

and

\lim_{y \to 1^-} \frac{\partial f}{\partial y} = 1

, the partial derivative is discontinuous along the line

y=1

. Since the initial condition

y(0)=1

lies on this line of discontinuity, the theorem does not guarantee a unique solution.

Question 4

For which of the following initial value problems does the Existence and Uniqueness Theorem fail to guarantee a unique solution in a neighborhood of the initial point?

$y' = \sqrt{x^2+y^2}, \; y(0)=0$ (correct answer)
$y' = x^2 + \arctan(y), \; y(1)=\pi$
$(y-1)y' = x, \; y(0)=0$
$y' = y \ln(x+1), \; y(1)=2$

Explanation: We analyze each case by checking the continuity of

f(x,y)

and

\frac{\partial f}{\partial y}

at the initial point

(x_0, y_0)

. A)

f(x,y) = \sqrt{x^2+y^2}

. This is continuous everywhere. However,

\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2+y^2}}

, which is undefined and thus not continuous at the initial point

(0,0)

. The theorem does not guarantee uniqueness. B)

f(x,y) = x^2 + \arctan(y)

and

\frac{\partial f}{\partial y} = \frac{1}{1+y^2}

are both continuous everywhere. A unique solution is guaranteed. C) Rewriting gives

y' = \frac{x}{y-1}

. Here

f(x,y)=\frac{x}{y-1}

and

\frac{\partial f}{\partial y} = -\frac{x}{(y-1)^2}

. At the initial point

(0,0)

, both are continuous. A unique solution is guaranteed. D)

f(x,y) = y \ln(x+1)

and

\frac{\partial f}{\partial y} = \ln(x+1)

are both continuous in a neighborhood of

(1,2)

(specifically for

x>-1

). A unique solution is guaranteed.

Question 5

Let $R$ be the largest open rectangular region in the $xy$ -plane containing the point $(1, 5)$ for which the Existence and Uniqueness theorem guarantees a unique solution to the IVP $y' = \frac{\sqrt{y-4}}{x^2-9}$ , $y(1)=5$ . Which of the following points lies on the boundary of $R$ ?

$(-3, 5)$ (correct answer)
$(0, 0)$
$(1, 4.1)$
$(2, 10)$

Explanation: Let

f(x,y) = \frac{\sqrt{y-4}}{x^2-9}

. For existence, we need

y-4 \ge 0 \implies y \ge 4

and

x^2-9 \neq 0 \implies x \neq \pm 3

. For uniqueness, we check

\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y-4}(x^2-9)}

. For this to be continuous, we need a stricter condition on

y

y-4 > 0 \implies y > 4

, and also

x \neq \pm 3

. The initial point is

(1,5)

. The region where the conditions for a unique solution hold is defined by

y>4

and

x \neq \pm 3

. The largest connected open region containing

(1,5)

is the set of points

(x,y)

such that

-3 < x < 3

and

y > 4

. This is an open rectangular strip. Its boundaries are the lines

x=-3

x=3

, and

y=4

. From the choices, the point

(-3, 5)

lies on the boundary line

x=-3

Question 6

For the linear initial value problem $(x-3)\ln(x) y' + \frac{1}{x-5} y = \sqrt{x+1}$ , $y(4)=2$ , what is the largest open interval on which a unique solution is guaranteed to exist?

$(3, \infty)$
$(0, \infty)$
$(1, 5)$
$(3, 5)$ (correct answer)

Explanation: When analyzing where a linear first-order differential equation has a unique solution, you need to identify where both the coefficient of

y'

and the functions in standard form are continuous and well-defined. First, rewrite the equation in standard form by dividing by the coefficient of

y'

y' + \frac{1}{(x-3)\ln(x)(x-5)} y = \frac{\sqrt{x+1}}{(x-3)\ln(x)}

For a unique solution to exist, both the coefficient of

y

and the right-hand side must be continuous. This requires examining where each component could fail:

$\sqrt{x+1}$ requires $x \geq -1$
$\ln(x)$ requires $x > 0$ and $\ln(x) \neq 0$ , so $x \neq 1$
The denominators cannot be zero: $x \neq 3$ and $x \neq 5$

Since

\ln(x) = 0

when

x = 1

, we have discontinuities at

x = 1, 3,

and

5

. The initial condition

y(4) = 2

tells us we need an interval containing

x = 4

. The real line is divided into intervals:

(-1,0)

(0,1)

(1,3)

(3,5)

, and

(5,\infty)

. Since

x = 4

lies in

(3,5)

, this is our guaranteed interval. Answer choice A

(3,\infty)

crosses the discontinuity at

x = 5

. Choice B

(0,\infty)

crosses discontinuities at

x = 1, 3,

and

5

. Choice C

(1,5)

crosses the discontinuity at

x = 3

. Study tip: Always identify all discontinuities first, then find the largest interval containing your initial condition that doesn't cross any discontinuity.

Question 7

Let $y' = f(x,y)$ be a differential equation. Suppose $f(x,y)$ is continuous on the entire $xy$ -plane, but its partial derivative $\frac{\partial f}{\partial y}$ is continuous everywhere except at a single point $(x_0, y_0)$ . Which of the following statements must be true for the initial value problem with the condition $y(x_0) = y_0$ ?

The theorem provides no conclusion about either existence or uniqueness of any solution.
A solution is guaranteed to exist, and it is guaranteed to be non-unique due to the discontinuity.
A unique solution is guaranteed to exist because the discontinuity occurs at only a single point.
A solution is guaranteed to exist, but the theorem provides no conclusion about its uniqueness. (correct answer)

Explanation: When you encounter questions about existence and uniqueness of solutions to differential equations, you need to recall the Existence and Uniqueness Theorem (Picard's Theorem). This theorem has two parts: if

f(x,y)

is continuous in a region containing

(x_0, y_0)

, then a solution exists; if additionally

\frac{\partial f}{\partial y}

is continuous in that region, then the solution is unique. In this problem,

f(x,y)

is continuous everywhere, so the existence condition is satisfied globally. Even though

\frac{\partial f}{\partial y}

has a discontinuity at

(x_0, y_0)

, the continuity of

f

itself guarantees that a solution exists. However, since

\frac{\partial f}{\partial y}

is discontinuous exactly at our initial point, the uniqueness condition fails, so the theorem cannot guarantee uniqueness. Looking at the wrong answers: A) incorrectly suggests we get no information about existence, but continuity of

f

does guarantee existence. B) makes the error of claiming non-uniqueness is guaranteed - while uniqueness isn't guaranteed by the theorem, this doesn't mean multiple solutions definitely exist. C) wrongly assumes that a single point of discontinuity doesn't matter for uniqueness, but any discontinuity of

\frac{\partial f}{\partial y}

at the initial point prevents the theorem from guaranteeing uniqueness. The correct answer is D: existence is guaranteed by continuity of

f

, but uniqueness cannot be concluded due to the discontinuity of

\frac{\partial f}{\partial y}

. Remember: existence requires continuity of

f

; uniqueness requires continuity of both

f

and

\frac{\partial f}{\partial y}

at the initial point.

Question 8

Consider the IVP $y' = 1 + y^2$ , $y(0) = 0$ . The functions $f(x,y)=1+y^2$ and $\frac{\partial f}{\partial y}=2y$ are continuous on the entire $xy$ -plane. Let $y(x)$ be the unique solution to this IVP. Which statement is true about the domain of this solution?

The solution $y(x)$ is unique and exists for all real numbers $x$ .
The solution $y(x)$ is unique but exists only on a finite interval containing $x=0$ . (correct answer)
The Existence and Uniqueness Theorem does not apply because $f(x,y)$ is nonlinear.
Multiple solutions exist for this IVP, so its domain cannot be uniquely determined.

Explanation: The Existence and Uniqueness Theorem guarantees a unique solution exists in some local neighborhood of the initial point. The fact that the conditions hold everywhere on the plane does not guarantee a global solution that exists for all

x \in \mathbb{R}

. To determine the actual interval of existence, we solve the IVP. It is separable:

\frac{dy}{1+y^2} = dx

. Integrating gives

\arctan(y) = x + C

. The initial condition

y(0)=0

implies

C=0

. The solution is

y(x) = \tan(x)

. The branch of the tangent function that passes through the origin is defined on the interval

(-\pi/2, \pi/2)

. The solution has vertical asymptotes at

x=\pm\pi/2

and thus 'blows up' in finite time. Therefore, the solution exists only on a finite interval, not for all real numbers.

Question 9

For the initial value problem defined by the implicit differential equation $e^{y} y' - x\sqrt{y} = 0$ , $y(1) = 1$ , which of the following statements is correct according to the Existence and Uniqueness Theorem?

The theorem does not guarantee uniqueness because the partial derivative with respect to $y$ is undefined at $y=0$ .
The theorem does not guarantee existence because $\sqrt{y}$ requires $y \ge 0$ .
A unique solution is guaranteed to exist in a neighborhood of $x=1$ . (correct answer)
The theorem cannot be applied because the differential equation is given in implicit form.

Explanation: When analyzing existence and uniqueness for differential equations, you need to check whether the Existence and Uniqueness Theorem's conditions are satisfied at your initial point and in its neighborhood. First, rewrite the equation in standard form:

y' = \frac{x\sqrt{y}}{e^y}

Now identify

f(x,y) = \frac{x\sqrt{y}}{e^y}

and check the theorem's requirements near the initial condition

(1,1)

. For existence, you need

f(x,y)

to be continuous near

(1,1)

. Since

e^y > 0

always and

\sqrt{y}

is continuous for

y > 0

, the function

f

is continuous in a neighborhood around

(1,1)

where

y > 0

. For uniqueness, you need

\frac{\partial f}{\partial y}

to be continuous near

(1,1)

. Computing this partial derivative:

\frac{\partial f}{\partial y} = \frac{x}{e^y}\left(\frac{1}{2\sqrt{y}} - \sqrt{y}\right)

. This is continuous near

(1,1)

since

y = 1 > 0

. Both conditions are satisfied, so C is correct - a unique solution exists in a neighborhood of

x = 1

. A is wrong because the partial derivative is well-defined at our initial point

y = 1

, not

y = 0

. B is incorrect because

\sqrt{y}

is perfectly fine at

y = 1

; we only need continuity near the initial condition, not globally. D misunderstands the theorem - you can always rewrite implicit equations in explicit form to apply the theorem. Study tip: Always check the theorem's conditions specifically at your initial point and its immediate neighborhood, not globally across the entire domain.

Question 10

Consider the IVP $y' = (y^2-1)^{1/3}$ , $y(3)=1$ . Which statement is correct based on the Existence and Uniqueness Theorem?

Both existence and uniqueness of a solution are guaranteed.
Existence of a solution is guaranteed, but uniqueness is not. (correct answer)
Existence of a solution is not guaranteed because the derivative of $f$ is undefined.
Neither existence nor uniqueness is guaranteed by the theorem.

Explanation: Let

f(x,y) = (y^2-1)^{1/3}

. The cube root function is continuous for all real inputs, and

y^2-1

is a polynomial, so their composition

f(x,y)

is continuous on the entire

xy

-plane. Since

f

is continuous at the initial point

(3,1)

, the theorem guarantees that a solution exists. For uniqueness, we examine

\frac{\partial f}{\partial y} = \frac{1}{3}(y^2-1)^{-2/3}(2y) = \frac{2y}{3(y^2-1)^{2/3}}

. The denominator is zero when

y^2-1=0

, i.e., at

y=\pm 1

. The partial derivative is not continuous at these values. Since the initial condition is

y(3)=1

, the uniqueness condition of the theorem is not met at the initial point. Therefore, existence is guaranteed, but uniqueness is not.