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Find a first order differential equation for the given family of curves. $$ y=x^{1 / 2}+c x $$

Short Answer

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Question: Find the first order differential equation for the given family of curves: \(y = x^{1/2} + cx\). Answer: \(y = x\frac{dy}{dx} - \frac{1}{2}x^{3/2} + x^{1/2}\).

Step by step solution

01

Differentiate with respect to x

First, we differentiate the given equation, y = x^(1/2) + cx, with respect to x using the power rule. $$ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} + c $$
02

Solve for c

Next, solve for the arbitrary constant c in terms of y, x, and dy/dx. $$ c = \frac{dy}{dx} - \frac{1}{2}x^{-1/2} $$
03

Substitute c back into the original equation

Now, we will substitute the expression for c in terms of y, x, and dy/dx back into the original equation $$ y = x^{1 / 2} + c x $$ Replace c with the expression we found in Step 2: $$ y = x^{1 / 2} + (\frac{dy}{dx} - \frac{1}{2}x^{-1/2})x $$
04

Simplify the equation

Simplify the equation and collect the terms to get the first order differential equation. $$ y - x^{1/2} = (\frac{dy}{dx} - \frac{1}{2}x^{-1/2})x $$ Move the term with x^(1/2) to the right side of the equality: $$ y = (\frac{dy}{dx} - \frac{1}{2}x^{-1/2})x + x^{1/2} $$ Expand the product: $$ y = x\frac{dy}{dx} - \frac{1}{2}x^{1/2}x + x^{1/2} $$ Simplify the term with x^(1/2): $$ y = x\frac{dy}{dx} - \frac{1}{2}x^{3/2} + x^{1/2} $$ So, the first order differential equation for the given family of curves is: $$ y = x\frac{dy}{dx} - \frac{1}{2}x^{3/2} + x^{1/2} $$

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Key Concepts

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

Differential Calculus
Differential calculus is a subfield of calculus concerned with the study of the rates at which quantities change. It is the mathematical foundation for understanding how variables evolve over time and is used to solve problems involving change and motion. Specifically, differential calculus deals with the concept of the derivative, which represents the rate of change of a function with respect to an independent variable.

For instance, if one needs to model the growth rate of a plant over time, differential calculus allows for calculating the speed at which the plant's height increases per unit of time. This real-world scenario is mirrored by a mathematical function where the height of the plant is a function of time. In the exercise given, the rate at which the function changes is captured by the first order differential equation, which is solved step by step through standard calculus techniques.
Power Rule Differentiation
The power rule is a basic rule of differentiation used in calculus to easily compute the derivative of a power function of the form \( f(x) = x^n \), where \( n \) is any real number. According to the power rule, the derivative of \( f(x) \) is \( f'(x) = nx^{n-1} \). This rule greatly simplifies the process of differentiation, especially when dealing with polynomial functions.

Applied to the textbook exercise, when the family of curves \( y = x^{1/2} + cx \) is differentiated with respect to \( x \), the power rule allows us to quickly find \( \frac{dy}{dx} = \frac{1}{2}x^{-1/2} + c \). Understanding and correctly applying the power rule is crucial for students as it underpins many other concepts and problem-solving techniques in differential calculus.
Arbitrary Constants in Differential Equations
In differential equations, an arbitrary constant represents a constant value that can take on any number within a set of possible values, typically real numbers. These constants arise naturally when integrating differential equations, embodying the general solution to a family of curves or functions rather than a unique solution.

Whenever a differential equation involves an arbitrary constant, it indicates there are infinitely many solutions, each corresponding to a different initial condition or path taken by a system. In our exercise, the constant \( c \) serves this purpose, representing a different curve within the family. Solving for \( c \) explicitly in terms of \( y \) and \( x \) illustrates how a particular solution can be obtained for a specific initial condition. This concept is fundamental for understanding the general behavior of dynamic systems and is a cornerstone of both pure and applied mathematics.

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Most popular questions from this chapter

Tanks \(T_{1}\) and \(T_{2}\) have capacities \(W_{1}\) and \(W_{2}\) liters, respectively. Initially they are both full of dye solutions with concentrations \(c_{1}\) and \(c_{2}\) grams per liter. Starting at \(t_{0}=0,\) the solution from \(T_{1}\) is pumped into \(T_{2}\) at a rate of \(r\) liters per minute, and the solution from \(T_{2}\) is pumped into \(T_{1}\) at the same rate. (a) Find the concentrations \(c_{1}(t)\) and \(c_{2}(t)\) of the dye in \(T_{1}\) and \(T_{2}\) for \(t>0\). (b) Find \(\lim _{t \rightarrow \infty} c_{1}(t)\) and \(\lim _{t \rightarrow \infty} c_{2}(t)\).

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In Exercises \(1-4\) find the equations of the trajectories of the given undamped equation. Identify the equilibrium solutions, determine whether they are stable or unstable, and plot some trajectories. $$ \mathrm{C} / \mathrm{G} \quad y^{\prime \prime}+y e^{-y}=0 $$

An infinite sequence of identical tanks \(T_{1}, T_{2}, \ldots, T_{n}, \ldots,\) initially contain \(W\) gallons each of pure water. They are hooked together so that fluid drains from \(T_{n}\) into \(T_{n+1}(n=1,2, \cdots) .\) A salt solution is circulated through the tanks so that it enters and leaves each tank at the constant rate of \(r \mathrm{gal} / \mathrm{min} .\) The solution has a concentration of \(c\) pounds of salt per gallon when it enters \(T_{1}\). (a) Find the concentration \(c_{n}(t)\) in \(\operatorname{tank} T_{n}\) for \(t>0\). (b) Find \(\lim _{t \rightarrow \infty} c_{n}(t)\) for each \(n\).

In connection with Eqn \((4.4 .15),\) suppose \(y(0)=0\) and \(y^{\prime}(0)=v_{0},\) where \(0

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