Question

The initial value problem \(y^{\prime}+y=q(t) y^{2}, y(0)=y_{0}\) is known to have solution $$ y(t)=\frac{3}{(1-3 t) e^{t}} $$ on the interval \(-\infty

Short answer

Question: Determine the coefficient function q(t) and the initial value y_0 of the given solution y(t) = 3/((1-3t)e^t) for the Riccati equation y'(t) + y(t) = q(t)y^2(t). Answer: The coefficient function is q(t) = (t^2e^t - t + 2/3)e^t and the initial value is y_0 = 3.

Step-by-step solution

Differentiate the given solution

Differentiate the given solution \(y(t)\) with respect to \(t\). Using the quotient rule, we get: $$ y^\prime(t) = \frac{-3(1-3t)e^t - 3(1-3t)e^t (-1)}{((1-3t)e^t)^2} $$ Simplify the expression: $$ y^\prime(t)= \frac{-6+9t}{((1-3t)e^t)^2} $$

Substitute \(y(t)\) and \(y^\prime(t)\) into the equation

Now substitute \(y(t)\) and \(y^\prime(t)\) into the given equation \(y^\prime+y=q(t)y^2\): $$ \frac{-6+9t}{((1-3t)e^t)^2} + \frac{3}{(1-3t)e^t} = q(t) \left(\frac{3}{(1-3t)e^t}\right)^2 $$

Solve for \(q(t)\)

.multiply both sides by \((1-3t)^2e^{2t}\) to get rid of the denominators: $$ (-6+9t)(1-3t)e^{t} + 3(1-3t)^2e^{2t} = 9q(t) $$ Let's simplify the equation: $$ (9t^2e^t - 9t + 6)e^t = 9q(t) $$ Divide both sides by 9: $$ q(t) = (t^2e^t - t + \frac{2}{3})e^t $$

Find the initial value \(y_0\)

Now, find the initial value \(y_0\) by evaluating the given solution \(y(t)\) at \(t=0\): $$ y_0 = y(0) = \frac{3}{(1 - 3\cdot 0)e^0} = \frac{3}{1} = 3 $$ So the coefficient function is \(q(t) = (t^2e^t - t + \frac{2}{3})e^t\) and the initial value is \(y_0 = 3\).

Key concepts

Initial Value Problem

An initial value problem in differential equations includes a differential equation along with a specified value, known as the initial value, at a particular point. These problems have great importance in understanding how functions evolve over time from a known starting condition. In our example, the differential equation is \(y^{\prime} + y = q(t) y^2\) complementing with \(y(0) = y_0\) as the initial condition. This means we need to determine \(y(t)\) from its derivative \(y^{\prime}\) while satisfying the condition where \(t=0\), \(y(t)\) equals \(y_0\).

Initial value problems allow us to track systems' behavior under certain initial conditions and predict future states. In our exercise, the specific solution \(y(t) = \frac{3}{(1-3t)e^t}\) satisfies both the differential equation and the initial condition, enabling us to find \(q(t)\) and \(y_0\). In practical terms, IPVs are crucial for modeling phenomena such as population dynamics, financial markets, and physical systems.

Quotient Rule

When dealing with differential equations, differentiation of functions may involve applying special techniques. The quotient rule is key to differentiating a function that is the division of two other functions. If you have a function \(f(t) = \frac{u(t)}{v(t)}\), the derivative \(f^{\prime}(t)\) is calculated as:\[\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v \cdot u^{\prime} - u \cdot v^{\prime}}{v^2}\]
This rule helps find the derivative of the solution \(y(t)\), which is more complex due to its fraction structure. In our problem, we used this rule to obtain \(y^{\prime}(t)\) from \(y(t) = \frac{3}{(1-3t)e^t}\). As we can see, the numerator is derived from multiplying the derivative of the numerator by the denominator, minus the numerator times the derivative of the denominator.
Understanding the quotient rule is foundational for tackling more involved calculus problems, especially those involving rational functions in both real and applied contexts.

Coefficient Function

A coefficient function defines how the variable part of a differential equation is scaled or transformed by a function. In our differential equation \(y^{\prime} + y = q(t) y^2\), \(q(t)\) represents such a function that modifies the rate of change as dictated by the equation. This function is typically dependent on the independent variable, \(t\), and can greatly influence the system's dynamics.

From the problem, after evaluating the expressions and simplifying, we derived \(q(t) = (t^2 e^t - t + \frac{2}{3}) e^t\). This coefficient function allows us to understand how the system interacts and affects the solution's behavior over different intervals. It acts as a parameter that shapes the trajectory of solutions, often informing predictions based on particular models like economic growth rates or population dynamics.

• Influences amplitude or rate of change in the solution.
• Dependent on the variable of interest.
• Shapes the outcome significantly.

Interval of Solution

The interval of solution in differential equations denotes the range of variable values over which a solution is valid and meaningful. In our exercise, the solution \(y(t) = \frac{3}{(1-3t)e^t}\) is defined on the interval \(-\infty < t < \frac{1}{3}\). This interval is crucial since it informs where the solution applies and is physically feasible.
\(t < \frac{1}{3}\) ensures that the denominator of the solution does not become zero, which would make the function undefined. Specifying these intervals helps us avoid singularities or points where solutions might behave irregularly.

Interpreting intervals is a key aspect of mathematical modeling and real-world applications, as they delineate boundaries, ensure solutions' validity, and guide parameter choices within those bounds. It ensures that solutions align with specific conditions and restrictions in practice.