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As stated in this section, the separation of equation (2) on page 42 requires division by \( p(y) \), and this may disguise the fact that the roots of the equation \( p(y)=0 \) are actually constant solutions to the differential equation. (a) To explore this further, separate the equation $$ \frac{d y}{d x}=(x-3)(y+1)^{2 / 3} $$ to derive the solution, $$ y=-1+\left(x^{2} / 6-x+C\right)^{3} $$ (b) Show that \( y \equiv-1 \) satisfies the original equation \( d y / d x=(x-3)(y+1)^{2 / 3} \) (c) Show that there is no choice of the constant \( C \) that will make the solution in part (a) yield the solution \( y \equiv-1 \) Thus, we lost the solution \( y=-1 \) when we divided by \((y+1)^{2 / 3}\).

Step by step solution

01

Separate the equation

First, we need to separate the variables. The given differential equation is \(\frac{dy}{dx} = (x-3)(y+1)^{2/3}\). To separate, we consider \(dy/(y+1)^{2/3} = (x-3)dx\).

02

Integrate the equation

Next, integrate both sides with respect to their respective variables. This results in \(\int dy/(y+1)^{2/3} = \int (x-3) dx\). Solving the integrals ends up with \(3(y+1)^{1/3} = x^2/2 - 3x + C.\)

03

Derive the solution

After the integration, arrange the solution in the form of \(y\). That is, \(y = -1 + (x^2/6 -x +C)^3\).

04

Verify that \(y \equiv -1\)

substituting \(y \equiv -1\) into the original equation, we get \(dy/dx = (x-3)(-1+1)^{2/3}\) which simplifies to 0 = 0, indicating that \(y \equiv -1\) is indeed a solution.

05

Realize the lost solution

Lastly, the solution that we found could not be made to fit \(y \equiv -1\) regardless of the constant \(C\) chosen. This highlights the lost solution when we divided by \((y+1)^{2/3}\)

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Key Concepts

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

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. They are used to describe a variety of phenomena such as movement, growth, and decay across different fields like physics, biology, and engineering.
Their structure often involves derivatives, which are rates of change, helping us understand dynamic systems.

• The simplest form is an ordinary differential equation which involves functions of one variable and their derivatives.
• Partial differential equations deal with functions of multiple variables.
• Solving these equations involves finding a function or a set of functions that satisfy the given relationship.

In our exercise, we started with the differential equation \( \frac{dy}{dx} = (x-3)(y+1)^{2/3} \), which means we're looking at how the variable \( y \) changes with respect to \( x \).
Solving such equations often requires techniques like separation of variables, which we'll explore a bit further.

Constant Solutions

Constant solutions are specific solutions to differential equations where the function's value does not change. This means the derivative of the function is zero.
In other words, constant solutions occur when the rate of change of the function is not affected by any variable changes.

• They are crucial in identifying stability within systems.
• In our problem, \( y \equiv -1 \) represents a constant solution, implying no change in the \( y \) value as \( x \) varies.
• To verify, substitute \( y = -1 \) back into the original differential equation. If both sides simplify to 0, it confirms the solution is constant.

Constant solutions can be easily overlooked during algebraic manipulation but serve as important solutions providing insights into the system behavior, such as equilibria in physical systems.

Integration

Integration is the mathematical process of finding the antiderivative, which is essentially the reverse of taking a derivative.
It plays a key role in solving differential equations by allowing us to find functions given their rates of change.

• When dealing with separable differential equations, we often use integration to find \( y \) in terms of \( x \).
• In our step-by-step solution, we integrated both sides of the equation \( \int \frac{dy}{(y+1)^{2/3}} = \int (x-3) dx \), resulting in a new relationship involving \( y \) and \( x \).
• After integration, rearranging gives us our solution: \( y = -1 + (x^2/6 - x + C)^3 \), where \( C \) is the constant of integration.

Careful integration helps us derive the general solution and often reveals information about the constants that particularize specific solutions, yet it can also lead to challenges, including potential lost solutions.

Lost Solutions in Separation of Variables

Using separation of variables is a common method for solving differential equations.
However, it involves isolating each variable on a different side of the equation before integrating, which might sometimes overlook important solutions.

• This occurs particularly when you divide by an expression that could equal zero. Doing so may eliminate valid solutions from consideration.
• In our exercise, the division by \( (y+1)^{2/3} \) caused us to ignore the solution \( y = -1 \).
• This is because when \( y+1 \) equates to zero, division results in an undefined state, potentially masking certain solutions.
• Thus, it's pivotal to consider these points where the denominator could be zero separately to ensure no solutions are lost.

Being mindful of these challenges prevents misunderstandings and ensures a more comprehensive grasp of the problem's solution space.