Question

(a) Solve the exact equation $$ \left(x^{2}+y^{2}\right) d x+2 x y d y=0 $$ implicitly, (b) For what choices of \(\left(x_{0}, y_{0}\right)\) does Theorem 2.3 .1 imply that the initial value problem $$ \left(x^{2}+y^{2}\right) d x+2 x y d y=0, \quad y\left(x_{0}\right)=y_{0}, $$ has a unique solution \(y=y(x)\) on some open interval \((a, b)\) that contains \(x_{0} ?\) (c) Plot a direction field and some integral curves for (A). From the plot determine, the interval \((a, b)\) of (b), the monotonicity properties (if any) of the solution of (B), and \(\lim _{x \rightarrow a+} y(x)\) and \(\lim _{x \rightarrow b-} y(x) .\) HiNT: Your answers will depend upon which quadrant contains \(\left(x_{0}, y_{0}\right)\).

Short answer

For a given initial point \((x_0, y_0)\), the general solution is: $$ \frac{1}{3}x^3 + x y^2 = C $$ The interval \((a, b)\), monotonicity properties of the solution, and the limits will depend on the quadrant in which \((x_0, y_0)\) lies: 1. If \((x_0, y_0)\) is in the first quadrant (\(x_0 > 0\), \(y_0 > 0\)): - The solution will be increasing as \(x\) increases. - The interval \((a, b)\) will be \((0, \infty)\). - \(\lim_{x \rightarrow 0+} y(x) = 0\) and \(\lim_{x \rightarrow \infty} y(x) = \infty\). 2. If \((x_0, y_0)\) is in the second quadrant (\(x_0 < 0\), \(y_0 > 0\)): - The solution will be decreasing as \(x\) increases. - The interval \((a, b)\) will be \((-\infty, 0)\). - \(\lim_{x \rightarrow -\infty} y(x) = \infty\) and \(\lim_{x \rightarrow 0-} y(x) = 0\). 3. If \((x_0, y_0)\) is in the third quadrant (\(x_0 < 0\), \(y_0 < 0\)): - The solution will be increasing as \(x\) increases. - The interval \((a, b)\) will be \((-\infty, 0)\). - \(\lim_{x \rightarrow -\infty} y(x) = -\infty\) and \(\lim_{x \rightarrow 0-} y(x) = 0\). 4. If \((x_0, y_0)\) is in the fourth quadrant (\(x_0 > 0\), \(y_0 < 0\)): - The solution will be decreasing as \(x\) increases. - The interval \((a, b)\) will be \((0, \infty)\). - \(\lim_{x \rightarrow 0+} y(x) = 0\) and \(\lim_{x \rightarrow \infty} y(x) = -\infty\).

Step-by-step solution

Convert the given equation into an exact differential equation

Rewrite the given equation as: $$ M(x, y)dx + N(x, y)dy = 0 $$ with \(M(x, y) = (x^2 + y^2)\) and \(N(x, y) = 2xy\).

Check if the equation is exact

To check if the equation is exact, we can verify that \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). If it's true, then our equation is exact. $$ \frac{\partial M}{\partial y} = 2y, \quad \frac{\partial N}{\partial x} = 2y $$ Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the given equation is exact.

Integrate to find the general solution

Integrate \(M(x, y)\) with respect to \(x\) and \(N(x, y)\) with respect to \(y\). Then equate the two resulting expressions. $$ \int M(x, y) dx = \int (x^2 + y^2) dx = \frac{1}{3}x^3 + x y^2 + g(y) $$ $$ \int N(x, y) dy = \int 2xy dy = x y^2 + h(x) $$ Now, equate the two integrals: $$ \frac{1}{3}x^3 + x y^2 + g(y) = x y^2 + h(x) $$ In this case, \(g(y) = 0\) and \(h(x) = \frac{1}{3}x^3\). Thus, the general solution is: $$ \frac{1}{3}x^3 + x y^2 = C $$

Determine the initial value problem uniqueness

According to Theorem 2.3.1, the initial value problem will have a unique solution if \(\frac{\partial M}{\partial x} \neq 0\) or \(\frac{\partial N}{\partial y} \neq 0\). Since \(\frac{\partial M}{\partial x} = 2x \neq 0\) and \(\frac{\partial N}{\partial y} = 2x \neq 0\), therefore the initial value problem has a unique solution \(y=y(x)\) on some open interval \((a,b)\) that contains \(x_0\), for \((x_0, y_0) \neq (0, 0)\).

Plot the direction field and integral curves

Analyzing the direction field and integral curves is not possible here, but you can use software like Matlab or Python(SciPy or NumPy) to plot the direction field. Based on the quadrant in which the point \((x_0, y_0)\) lies, the interval \((a, b)\) in (b) will change, as well as the monotonicity properties of the solution of (B), and the limits \(\lim_{x \rightarrow a+} y(x)\) and \(\lim_{x \rightarrow b-} y(x)\).

Key concepts

Initial Value Problem

An Initial Value Problem in differential equations is where you're given a differential equation and initial condition(s) that the solution must satisfy. The general form of an initial value problem can be written as:\[ F(x, y, y') = 0, \quad y(x_0) = y_0 \]In this form, you're asked to find a function \( y = y(x) \) that not only satisfies the differential equation internally but also the specific initial condition. This ensures that you're pinpointing one specific solution out of potentially many that merely solve the differential equation. For our specific exercise, the initial value problem is given as:\[ (x^2+y^2) dx + 2xy dy = 0, \quad y(x_0) = y_0 \]This condition \( y(x_0) = y_0 \) sets the stage for finding a unique trajectory or curve that the solution will follow based on the initial conditions provided.

Theorem 2.3.1

Theorem 2.3.1 is a fundamental result pertaining to the existence and uniqueness of solutions for differential equations. This theorem essentially provides conditions under which a definite unique solution exists for a differential equation with an initial condition. For an exact differential equation,- if the partial derivatives satisfy certain criteria, such as: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) and at least one other condition related to partial derivatives, it fosters a unique solution.In the context of our problem, Theorem 2.3.1 helps us ascertain when the initial value problem has a unique solution on an interval containing \( x_0 \). Specifically, it dictates that if \( \frac{\partial M}{\partial x} eq 0 \) or \( \frac{\partial N}{\partial y} eq 0 \), then there is a guarantee of uniqueness in the solution, barring the critical point \( (x_0, y_0) eq (0, 0) \).

Direction Field

A Direction Field (or slope field) is a visualization tool used to illustrate the solutions to a differential equation within a particular area. It’s like a map showing potential paths that a solution could take through the phase space. By plotting tiny line segments representing the slope given by the differential equation at various points, you get an overall glimpse of what the solution curves might look like.In part (c) of our exercise, you can plot a direction field for the differential equation using computational tools. Once plotted, the direction field enables you to identify the solution behaviors based on different initial conditions (initial values \( x_0, y_0 \)). Observing the direction field can also help infer the interval that contains \( x_0 \), and how the behavior of the solutions might change based on their monotonicity across the graph. This visualization effectively aids in understanding the trajectories that are possible given the constraints.

Unique Solution

A Unique Solution to an initial value problem or differential equation means there's exactly one function that satisfies both the differential equation and the initial conditions provided. The uniqueness is crucial because it assures that the path or trajectory is consistent and predictable from the conditions set initially.In the differential equation context we're dealing with, a unique solution is guaranteed by the conditions set in Theorem 2.3.1. In our exercise, uniqueness is validated by confirming that:- \( \frac{\partial M}{\partial x} = 2x eq 0 \)- \( \frac{\partial N}{\partial y} = 2x eq 0 \)These conditions ensure no ambiguity exists in the solution path, except for \( (x_0, y_0) = (0, 0) \). Thus, once an initial value is placed within any other point, the solution \( y = y(x) \) is determined precisely and does not waver from that path across the confirmed interval.