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Chapter 2: Problem 16

\(y^{\prime}=y+y^{-1}, \quad y(0)=-1\)

Short Answer

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Question: Given a first-order differential equation \(y^{\prime} = y + y^{-1}\) and an initial condition \(y(0) = -1\), find the function \(y(x)\) that satisfies this equation. Answer: The equation \(y^{\prime} = y + y^{-1}\) does not have a closed-form solution. An approximate solution can be found using numerical methods for the given initial condition, \(y(0) = -1\).

Step by step solution

01

Substitution

Let us introduce a substitution: \(u = y + y^{-1}\). Then, \(y = u - y^{-1}\). Differentiating both sides of \(y = u - y^{-1}\) with respect to \(x\), we get: \(y' = u' + y^{-2}y'\). Now substitute the expressions for \(y\) and \(y'\) into the given differential equation: \(u' + y^{-2}y' = u + y^{-1}\).
02

Solve for \(u'\)

Rearrange the equation to solve for \(u'\): \(u' = u + y^{-1} - y^{-2}y'\). Now, let's express \(y^{-1}\) and \(y^{-2}\) in terms of \(u\): \(y^{-1} = u - y\), and \(y^{-2} = (u - y)^{-2}\). Substitute into the above equation: \(u' = u + (u - y) - (u - y)^{-2}(u' + (u - y)^2)\).
03

Solve for \(y\) in terms of \(u\)

Simplify the equation and solve for \(y\): \(u' = u + u - y -u(u' + u - 2y)\), \(u' = 2u - y -u^2u' + 2uy\), \((1 - u^2)u' = 2u - y + 2uy\), \((1 - u^2)u' = 2u(1 + y)\). Now, let's rewrite the equation in terms of \(y\): \(y = (\frac{2u(1 + y)}{(1 - u^2)})'\).
04

Integrate both sides of the equation

Integrate both sides w.r.t \(x\): \(\int y dx = \int \frac{2u(1 + y)}{(1 - u^2)} dx\). To solve this integral, we will need to find the function \(u(x)\) that is a solution to the equation \((1 - u^2)u' = 2u(1 + y)\). Unfortunately, this equation is not solvable in a closed-form expression, so we cannot find an explicit solution \(y(x)\). However, we can solve this equation numerically for a given initial condition, \(y(0)=-1\), which could help us find an approximate solution for the function \(y(x)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is instrumental in simplifying differential equations that might otherwise be difficult to solve. By introducing a new variable to replace a complex expression, the equation becomes more manageable. This method often turns a tricky problem into one that we can solve with standard techniques. For example, in the exercise, the substitution \(u = y + y^{-1}\) simplifies the equation by reducing the original terms into something that can be differentiated and integrated more straightforwardly. Always remember, when substituting, we also need to change back to the original variable after solving the equation with the new one.

Solving Differential Equations
Differential equations describe how quantities change and are crucial in science and engineering. Solving these equations can provide us with functions that predict future events based on current values. There are several methods to solve differential equations, such as separation of variables, integrating factors, and characteristic equations. These methods require an understanding of calculus and, at times, some ingenuity. The key is to rewrite the equation in a form where we can apply known integration or differentiation techniques to find the solution.

Numerical Solutions of Differential Equations
When differential equations cannot be solved analytically, we resort to numerical methods to approximate solutions. Techniques like Euler's Method, the Runge-Kutta methods, and finite difference methods are powerful tools that help us estimate the solutions of differential equations by iterating through a series of points. Numerical solutions are especially useful in modeling real-world scenarios where exact solutions are neither feasible nor necessary. The exercise showcases a situation where a numerical approach might be the only way forward since analytic solution paths are not apparent.

Initial Value Problem
An initial value problem is a type of differential equation that comes with a specified value at a particular point, known as the initial condition. This condition allows us to find a unique solution that fits the scenario being modeled. The initial condition helps ensure that the approximate numerical solutions are tailored to the specific situation. For instance, in the exercise given, the condition \(y(0)=-1\) is essential for determining the unique behavior of the curve \(y(x)\) near \(x=0\). With this information, numerical methods can be employed to predict the behavior of \(y\) as \(x\) changes from the initial point.

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Most popular questions from this chapter

Use the ideas of Exercise 32 to solve the given initial value problem. Obtain an explicit solution if possible. $$ y^{\prime}=2 t+y+\frac{1}{2 t+y}, \quad y(1)=1 $$

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Consider the initial value problem $$ y^{\prime}=\sqrt{1-y^{2}}, \quad y(0)=0 . $$ (a) Show that \(y=\sin t\) is an explicit solution on the \(t\)-interval \(-\pi / 2 \leq t \leq \pi / 2\). (b) Show that \(y=\sin t\) is not a solution on either of the intervals \(-3 \pi / 2t^{*}\) ? (d) Show that the solution of the initial value problem is given by $$ y=\left\\{\begin{array}{lrl} -1, & -\infty &

A student performs the following experiment using two identical cups of water. One cup is removed from a refrigerator at \(34^{\circ} \mathrm{F}\) and allowed to warm in its surroundings to room temperature \(\left(72^{\circ} \mathrm{F}\right)\). A second cup is simultaneously taken from room temperature surroundings and placed in the refrigerator to cool. The time at which each cup of water reached a temperature of \(53^{\circ} \mathrm{F}\) is recorded. Are the two recorded times the same or not? Explain.
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