Question

Stars indicate the approximate level of difficulty, from easiest ( $\ast$ ) to most difficult $(\star \star \star )$. Warning: Even when the motion is nonchaotic, it can be very sensitive to tiny errors. In several of the computer problems you may need to increase your working precision to get satisfactory results. Consider the nonlinear first-order equation $\dot{x}=2\sqrt{x-1}$. (a) By separating variables, find a solution $x_1(t)$. (b) Your solution should contain one constant of integration $k$, so you might reasonably expect it to be the general solution. Show, however, that there is another solution, $x_2(t)=1,$ that is not of the form of $x_1(t)$ whatever the value of $k$. (c) Show that although $x_1(t)$ and $x_2(t)$ are solutions, neither $A{x}_1(t),$ nor $B{x}_2(t),$ nor $x_1(t)+x_2(t)$ are solutions. (That is, the superposition principle does not apply to this equation.)

Short answer

(a) $x_1(t)=(t+\frac{C}{2}{)}^2+1$; (b) $x_2(t)=1$; (c) Superposition does not apply.

Step-by-step solution

Separate Variables

Separate variables in the differential equation $\dot{x}=2\sqrt{x-1}$. This gives the form $\frac{dx}{\sqrt{x-1}}=2dt$.

Integrate Both Sides

Integrate both sides: $\int \frac{dx}{\sqrt{x-1}}=\int 2dt$. The left side integrates to $2\sqrt{x-1}$ and the right side to $2t+C$, where $C$ is a constant of integration. Thus, $2\sqrt{x-1}=2t+C$.

Solve for x

To express $x$ in terms of $t$, first divide both sides by 2: $\sqrt{x-1}=t+\frac{C}{2}$. Squaring both sides results in $x-1=(t+\frac{C}{2}{)}^2$. Add 1 to both sides to solve for $x$: $x_1(t)=(t+\frac{C}{2}{)}^2+1$.

Find Alternative Solution

Verify if $x_2(t)=1$ is a solution to the differential equation. Substitute $x_2(t)=1$ into $\dot{x}=2\sqrt{x-1}$: $\dot{x}=0$ and $2\sqrt{1-1}=0$. Since both sides are equal, $x_2(t)=1$ is indeed a solution.

Verify Non-linear Superposition

For non-linear equations, solutions $A{x}_1(t)$, $B{x}_2(t)$, or $x_1(t)+x_2(t)$ generally do not hold. Verify each case.

Test x1 + x2 for Superposition

Compute $x_1(t)+x_2(t)=(t+\frac{C}{2}{)}^2+2$ and substitute into $\dot{x}$. The equation $\dot{x}=2\sqrt{(t+\frac{C}{2}{)}^2+2-1}$ yields a different form, confirming that $x_1(t)+x_2(t)$ is not a solution.

Confirm Non-Applicability of Superposition

Repeat similar calculations for $A{x}_1(t)$ or $B{x}_2(t)$ by substituting them into the differential equation. In all traditional linear combinations, they should fail to satisfy the original differential equation $\dot{x}=2\sqrt{x-1}$. This confirms the non-linearity and invalidity of superposition for this problem.

Key concepts

Superposition Principle

In the context of differential equations, the superposition principle tells us that if we have two or more solutions to a linear differential equation, then any linear combination of them is also a solution. This principle is fundamental when dealing with linear systems, allowing solutions to be constructed from simpler components.

However, in the case of nonlinear differential equations, such as the one presented in this problem, the superposition principle does not apply. That is because the combination of solutions can lead to terms that do not satisfy the original nonlinear equation. This is why in this exercise, even though $x_1(t)$ and $x_2(t)$ are solutions, neither $A{x}_1(t)$ nor $B{x}_2(t)$ nor $x_1(t)+x_2(t)$ satisfy the differential equation $\dot{x}=2\sqrt{x-1}$. Nonlinear equations are picky; each solution needs to be verified independently.

Separation of Variables

Separation of variables is a powerful technique used to solve differential equations. It involves rearranging an equation so that each variable occurs on only one side of the equation. This method works particularly well for differential equations that can be expressed in a form where the function of $x$ is separated from the function of $t$.

In this exercise, the differential equation $\dot{x}=2\sqrt{x-1}$ was separated into $\frac{dx}{\sqrt{x-1}}=2dt$, enabling us to integrate each side independently. By performing the integrals, we solved for the function of $x$ in terms of $t$, thus arriving at the potential solution $x_1(t)$. An essential aspect of separation of variables is ensuring that all solutions fit within the functional constraints posed by the problem.

Constant of Integration

The constant of integration appears when we integrate a differential equation. It represents the family of solutions that stem from the addition of an arbitrary constant after an indefinite integral. This constant is crucial because it accounts for all possible solution shifts along the $t$-axis.

In the solution presented, after integrating both sides of the equation $\int \frac{dx}{\sqrt{x-1}}=\int 2dt$, we acquired an expression with the constant of integration $C$. Solving for $x$, we found $x_1(t)=(t+\frac{C}{2}{)}^2+1$. This constant $C$ is the bridge to different particular solutions which explain initial conditions or specific scenarios in the context of the problem.

First-Order Differential Equation

A first-order differential equation is an equation involving the first derivative of an unknown function. It provides valuable insights into the behavior of a system by defining the rate of change of a quantity over time.

The provided exercise dealt with the first-order differential equation $\dot{x}=2\sqrt{x-1}$. Here, $\dot{x}$ signifies the first derivative of $x$ with respect to time $t$. Solving such equations involves finding the function $x(t)$ that satisfies this relationship. By solving the equation using separation of variables, the students derived potential solutions. Understanding the nature and solutions of first-order differential equations is fundamental in many areas of engineering, physics, and beyond, as it helps describe systems' dynamics effectively.