Question

A function \(y(t)\) satisfies the differential equation \(\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{y}^{4}-6 \mathrm{y}^{3}+5 \mathrm{y}^{2}\) (a) What are the constant solutions of the equation? (b) For what values of \(y\) is y decreasing ?

Short answer

Question: List the constant solutions to the given differential equation and provide the range of values for y where the function is decreasing. Answer: The constant solutions to the given differential equation are y = 0, y = 1, and y = 5. The function is decreasing for values of y in the interval (1, 5).

Step-by-step solution

(a) Step 1: Write down the given differential equation

The given differential equation is \(\frac{\mathrm{dy}}{\mathrm{dt}}=y^{4}-6y^{3}+5y^{2}\).

(a) Step 2: Set the derivative equal to zero and solve for y

To find constant solutions, we need to find values of y for which the derivative is zero. So, we set the derivative equal to zero and solve for y: $$0=y^{4}-6y^{3}+5y^{2}$$

(a) Step 3: Factor the equation and find the constant values of y

We can factor the equation as follows: $$0=y^{2}(y^{2}-6y+5)$$ This gives us two factors: \(y^2\) and \(y^2-6y+5\). For \(y^2\), the values of y are 0. For \(y^2-6y+5\), we have \(y^2-6y+5 = (y-1)(y-5)\). The values of y are 1 and 5. The constant solutions of the equation are y = 0, y = 1, and y = 5.

(b) Step 1: Analyze the sign of the derivative

To determine when the function y is decreasing, we need to analyze when the derivative is negative. The function is decreasing if \(\frac{\mathrm{dy}}{\mathrm{dt}}<0\). Using the factored form of the differential equation, we have: $$\frac{\mathrm{dy}}{\mathrm{dt}}=y^{2}(y^{2}-6y+5)$$

(b) Step 2: Determine when the derivative is negative

The first term, \(y^2\), is non-negative for all values of y. The second term, \((y^2-6y+5)\), is the product of \((y-1)(y-5)\). Consider three cases: 1. When \(0 < y < 1\), both terms \((y-1)\) and \((y-5)\) are negative, making their product positive. Thus, the function is increasing in this interval. 2. When \(1 < y < 5\), \((y-1)\) is positive, and \((y-5)\) is negative, making their product negative. Thus, the function is decreasing in this interval. 3. When \(y > 5\), both terms \((y-1)\) and \((y-5)\) are positive, making their product positive. Thus, the function is increasing in this interval. The function is decreasing for values of \(y\) in the interval \((1, 5)\).

Key concepts

Constant Solutions in Differential Equations

When working with differential equations in calculus, a crucial aspect to understand is the concept of constant solutions. These are solutions where the derivative of the function is zero. In the context of our example, the differential equation has the form \[ \frac{\mathrm{dy}}{\mathrm{dt}}=y^{4}-6y^{3}+5y^{2} \.\]

To find the constant solutions, we set the derivative equal to zero because we're essentially looking for points where the function's rate of change—its growth or decay—stops, making it a flat line or a plateau on a graph.
Factoring allows us to simplify the complex expression into manageable parts to solve for values of \( y \). In this case, by factoring \( y^{2}(y^{2}-6y+5) \), we find that the constant solutions where the derivative equals zero are \( y = 0 \), \( y = 1 \), and \( y = 5 \).
Understanding that at these points the function doesn't increase or decrease is fundamental in further analyses—like when evaluating the behavior of a function within an interval or in optimization problems.

Derivatives and Their Role in Analyzing Functions

Derivatives are one of the most powerful tools in calculus for analyzing functions. In our example, the derivative is represented by \[ \frac{\mathrm{dy}}{\mathrm{dt}} \.\]

Derivatives measure how a function changes as its input changes, giving us the rate at which \( y \) value varies with respect to \( t \). This knowledge is vital when trying to determine when the function is increasing or decreasing. For instance, a positive derivative indicates an increasing function, while a negative derivative shows that the function is decreasing.
In our analysis, we use the sign of the derivative to determine that the function \( y(t) \) is decreasing when \( y \) is in the interval \((1, 5)\). The factored form of the derivative, \( y^{2}(y-1)(y-5) \), helped us conclude this behavior by examining the sign changes within the specified intervals, highlighting the derivative's role not only as a numerical measure but also as a signpost for function behavior.

Factoring Polynomials in Differential Equations

Factoring polynomials is a skill that turns intricate polynomial expressions into products of simpler expressions, simplifying our analysis significantly. In our differential equation, we were given \[ y^{4}-6y^{3}+5y^{2} \.\]

Factoring can unravel the structure of the equation, revealing crucial information about its solutions and their multiplicities. The factored form \( y^{2}(y^{2}-6y+5) \) breaks down the original polynomial into its component parts, where we can easily find the roots. After factoring further to \( y^{2}(y-1)(y-5) \), it became clear that \( y = 0 \), \( y = 1 \), and \( y = 5 \) are significant values where the function's behavior changes.
Additionally, these factored polynomials aid in identifying intervals where our function increases or decreases. Therefore, learning how to factor polynomials efficiently is an invaluable skill in solving and understanding differential equations and calculus as a whole.