Direction Fields & Isoclines - Differential Equations

Consider the differential equation $\frac{dy}{dx} = y^2 - 4y + 3$ . Based on the direction field for this equation, which statement best describes the long-term behavior of solutions with initial conditions $y(0) = 2.5$ ?
Answer: The solution approaches $y = 1$ as $x \to \infty$ and exhibits exponential growth away from $y = 3$ First, find equilibria by setting $y^2 - 4y + 3 = 0$ , which gives $(y-1)(y-3) = 0$ , so $y = 1$ and $y = 3$ . For $y = 2.5$ (between equilibria), $\frac{dy}{dx} = (2.5)^2 - 4(2.5) + 3 = 6.25 - 10 + 3 = -0.75 < 0$ . Since the derivative is negative, the solution decreases toward $y = 1$ . Testing stability: for $y$ slightly above 1, $\frac{dy}{dx} < 0$ ; for $y$ slightly below 3, $\frac{dy}{dx} < 0$ . Thus $y = 1$ is stable and $y = 3$ is unstable.
A student observes a direction field and correctly concludes that it must represent an autonomous differential equation of the form $y' = f(y)$ . Which of the following observations is sufficient to justify this conclusion?
Answer: All slope marks along any horizontal line are parallel. An autonomous differential equation has the form $y' = f(y)$ , which means the slope of the solution depends only on the value of $y$ , not on $t$ . Geometrically, this means that for any given height $y=c$ , the slope $f(c)$ is the same for all values of $t$ . Therefore, all the slope marks along any horizontal line $y=c$ must be parallel to each other. If slopes are parallel along vertical lines, the equation is of the form $y'=f(t)$ . Symmetry with respect to the origin is a different property not necessarily related to being autonomous.
A direction field for $\frac{dy}{dx} = g(x,y)$ exhibits the property that all isoclines with positive slope values are closed curves, while isoclines with negative slope values extend infinitely. Based on this information alone, which characteristic is most likely true about the function $g(x,y)$ ?
Answer: The function $g(x,y)$ has a local maximum at some point, creating bounded level sets for values near this maximum If isoclines with positive slope (where $g(x,y) = k > 0$ ) are closed curves, this suggests that positive values of $g$ are confined to bounded regions. This typically occurs when $g$ has a local maximum - the level sets $g(x,y) = k$ for $k$ near the maximum value form closed curves around the maximum point. For negative values ( $k < 0$ ), if these extend infinitely, it suggests $g$ can achieve arbitrarily negative values in unbounded regions. Choice B is less precise because it doesn't explain why positive values would be confined. Choices C and D don't directly explain the closed/open nature of the isoclines.
Consider the differential equation $y' = f(t,y)$ . Let $C$ be an isocline defined by the equation $g(t,y)=k$ corresponding to slope $m$ , so $f(t,y) = m$ for all $(t,y)$ on $C$ . Under which general condition is the isocline $C$ also a solution curve?
Answer: The isocline must be a line with slope $m$ . For an isocline to also be a solution curve, the slope of the isocline itself must be equal to the slope $m$ defined by the direction field at every point on the curve. If the isocline is a line $y = mt + b$ , its slope is $m$ everywhere. This matches the condition $y' = m$ for that isocline. Thus, the line is a solution. A horizontal line ( $m=0$ ) is a special case. An autonomous DE is not a sufficient condition. A nullcline ( $f(t,y)=0$ ) is an isocline for $m=0$ , but it is only a solution if it is a horizontal line (an equilibrium solution).
Consider solutions to the differential equation $y' = \cos(t) - y$ . Solution curves will have a local maximum or minimum at points $(t,y)$ that lie on which curve?
Answer: $y = \cos(t)$ A solution curve $y(t)$ has a local maximum or minimum when its derivative is zero, i.e., $y'(t)=0$ . This is a necessary condition from calculus. For the given differential equation, this means we must have $\cos(t) - y = 0$ . Solving for $y$ gives the curve $y = \cos(t)$ . This curve is the nullcline of the differential equation, which is the set of all points where solution curves have horizontal tangents.
For the differential equation $dy/dt = y^2 - t$ , find the equation of the curve in the $ty$ -plane along which the slopes of the solution curves are equal to the $y$ -coordinate of the point.
Answer: $t = y^2 - y$ The slope of a solution curve at a point $(t,y)$ is given by $dy/dt = y^2 - t$ . The condition specified in the problem is that this slope must be equal to the $y$ -coordinate, so we must have $dy/dt = y$ . Setting the two expressions for the slope equal gives the equation $y^2 - t = y$ . To express this as a curve, we can solve for $t$ : $t = y^2 - y$ . This is the equation of the curve where the condition holds. Choosing $y = \sqrt{t}$ corresponds to finding the nullcline ( $y'=0$ ), not where $y'=y$ .
Consider the differential equation $y' = \sqrt{t^2 + y^2} - 3$ . The isocline for a certain slope $c$ passes through the point $(4, -3)$ . What is the value of $c$ ?
Answer: $2$ An isocline is a curve where the slope $y'$ is constant. Let this constant slope be $c$ . The equation for the isocline is $c = \sqrt{t^2 + y^2} - 3$ . We are given that this curve passes through the point $(t,y) = (4, -3)$ . We can substitute these values into the isocline equation to find $c$ : $c = \sqrt{(4)^2 + (-3)^2} - 3 = \sqrt{16 + 9} - 3 = \sqrt{25} - 3 = 5 - 3 = 2$ .
For the differential equation $y' = y - t^2$ , in which region of the $ty$ -plane are the solution curves concave up?
Answer: The region where $y > t^2 + 2t$ . Concavity is determined by the sign of the second derivative, $y''$ . We first find $y''$ by differentiating $y'$ with respect to $t$ , using the chain rule: $y'' = \frac{d}{dt}(y - t^2) = \frac{dy}{dt} - 2t$ . Next, we substitute the original expression for $y'$ : $y'' = (y - t^2) - 2t = y - t^2 - 2t$ . Solution curves are concave up where $y'' > 0$ . Therefore, we need $y - t^2 - 2t > 0$ , which simplifies to $y > t^2 + 2t$ .
For the differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{x}{y}$ , a student attempts to analyze the direction field by finding isoclines. When looking for isoclines with slope $m = 2$ , which equation correctly represents these isoclines?
Answer: $\frac{y}{x} + \frac{x}{y} = 2$ , which simplifies to $x^2 + y^2 = 2xy$ and represents the line $y = x$ Setting $\frac{y}{x} + \frac{x}{y} = 2$ and multiplying by $xy$ : $y^2 + x^2 = 2xy$ . Rearranging: $x^2 - 2xy + y^2 = 0$ , which factors as $(x-y)^2 = 0$ . This gives $x = y$ , so the isocline is the line $y = x$ . We can verify: along $y = x$ , $\frac{dy}{dx} = \frac{x}{x} + \frac{x}{x} = 1 + 1 = 2$ . Choice A has an error in the algebra, Choice B incorrectly claims no real solutions, and Choice C stops before completing the factorization.
The isoclines for the differential equation $\frac{dy}{dx} = xy - y^2$ with slope $m = 0$ form curves in the xy-plane. Along these isoclines, which characteristic best describes the behavior of solution curves?
Answer: Solution curves have horizontal tangent lines and may represent local extrema of the solutions Isoclines with slope $m = 0$ are found by setting $xy - y^2 = 0$ , giving $y(x-y) = 0$ , so $y = 0$ or $y = x$ . Along these curves, $\frac{dy}{dx} = 0$ , meaning solution curves have horizontal tangent lines. These represent points where solutions have local maxima, minima, or inflection points. The direction field shows horizontal line segments along these isoclines.
Consider the differential equation $dy/dt = t/y$ . What is the geometric description of the isoclines for this equation?
Answer: A family of lines passing through the origin. Isoclines are curves where the slope $dy/dt$ is constant. Let $dy/dt = c$ for some constant $c$ . The equation for the isoclines is $t/y = c$ . This can be rewritten as $y = (1/c)t$ (for $c \neq 0$ ) or $t=0$ (for $c=0$ ). This equation describes the set of all lines passing through the origin with slope $1/c$ . The solution curves, $y^2 - t^2 = K$ , are hyperbolas, which is a common point of confusion.
Consider the direction field for $\frac{dy}{dx} = x^2 - y^2$ . A student incorrectly claims that since this equation is separable, all solution curves must be symmetric about the line $y = x$ . Which feature of the direction field most directly contradicts this claim?
Answer: The slopes at points $(a,b)$ and $(b,a)$ are generally different unless $a^2 - b^2 = b^2 - a^2$ When analyzing direction fields, you need to understand how the differential equation determines the slope at each point and what symmetry properties (if any) exist in the solution family. The key insight is to test the student's symmetry claim by examining what happens when you swap coordinates. If solution curves were symmetric about the line $y = x$ , then the slope at point $(a,b)$ should equal the slope at point $(b,a)$ . Let's check: at $(a,b)$ , the slope is $a^2 - b^2$ , while at $(b,a)$ , the slope is $b^2 - a^2 = -(a^2 - b^2)$ . These slopes are equal only when $a^2 - b^2 = 0$ , meaning $a = \pm b$ . For most points, the slopes differ, directly contradicting the symmetry claim. Answer choice A is incorrect because while $y = x$ is indeed a solution curve (since $x^2 - x^2 = 0$ ), this doesn't contradict symmetry—it would actually be expected for a symmetric family. Answer choice B is wrong because it supports rather than contradicts the student's claim, and the symmetric isoclines don't guarantee symmetric solutions. Answer choice D is irrelevant since exactness has nothing to do with this type of symmetry property. Study tip: When evaluating symmetry claims about solution families, always test specific points by plugging coordinates into the differential equation. Don't assume that separable equations or symmetric isoclines automatically produce symmetric solution curves—verify by checking the actual slope values.
For the autonomous equation $\frac{dy}{dx} = y(y-2)(y-4)$ , the direction field reveals several equilibrium solutions. A student observes that solutions starting near $y = 2$ appear to move away from this value. To verify this observation analytically, which approach most directly confirms the stability of $y = 2$ ?
Answer: Test the sign of $f(y) = y(y-2)(y-4)$ for values slightly above and below $y = 2$ When analyzing the stability of equilibrium points in autonomous differential equations, you need to understand how solutions behave near those equilibrium values. The most direct approach is examining the sign of the derivative function around the equilibrium point. For the equilibrium $y = 2$ , you want to test what happens to solutions that start slightly above or below this value. Since $\frac{dy}{dx} = f(y) = y(y-2)(y-4)$ , the sign of $f(y)$ tells you whether $y$ is increasing or decreasing. If you test $y = 1.9$ (slightly below 2), you get $f(1.9) = (1.9)(-0.1)(-2.1) > 0$ , so $\frac{dy}{dx} > 0$ and $y$ increases toward 2. If you test $y = 2.1$ (slightly above 2), you get $f(2.1) = (2.1)(0.1)(-1.9) < 0$ , so $\frac{dy}{dx} < 0$ and $y$ decreases toward 2. This confirms $y = 2$ is stable. Answer C correctly identifies this approach. Answer A calculates $f'(2)$ , which gives information about linearized stability but isn't the most direct method for understanding the actual behavior described by the student. Answer B involves solving the equation explicitly, which is unnecessarily complex and doesn't directly address local behavior near the equilibrium. Answer D examines the second derivative, which relates to concavity of the function $f(y)$ but doesn't directly determine stability. Study tip: For autonomous equations, always test the sign of $\frac{dy}{dx}$ on both sides of an equilibrium point to quickly determine whether solutions move toward or away from that point.

Start this quiz