Question

Verify that the given function or functions is a solution of the differential equation. $$ t y^{\prime}-y=t^{2} ; \quad y=3 t+t^{2} $$

Short answer

Question: Verify that the function \(y=3t+t^2\) is a solution to the differential equation \(ty'-y=t^2\). Answer: Using the steps of calculating the derivative of y with respect to t, plugging the function and its derivative into the differential equation, and simplifying the expression, we confirmed that both sides of the equation are equal. Therefore, the function \(y=3t+t^2\) is indeed a solution to the differential equation \(ty'-y=t^2\).

Step-by-step solution

Calculate the derivative of y with respect to t

Differentiate the given function \(y=3t+t^2\) with respect to t: $$ \begin{aligned} y^{\prime}=\frac{d y}{d t} &=\frac{d}{d t}(3 t+t^{2}) \\ &=3+2 t \end{aligned} $$

Plug the function and its derivative into the differential equation

Now, we will substitute the function and its derivative into the differential equation \(ty'-y=t^2\), confirming that it satisfies the equation. $$ \begin{aligned} t y^{\prime}-y &=t(3+2 t)-(3 t+t^{2}) \end{aligned} $$

Simplify the expression and check if both sides of the equation are equal

Simplify the expression on the left side of the equation and compare it to the right side of the equation, which is \(t^2\). $$ \begin{aligned} t y^{\prime}-y &=t(3+2 t)-(3 t+t^{2}) \\ &=3 t+2 t^{2}-3 t-t^{2} \\ &=t^{2} \end{aligned} $$ Since both sides of the equation are equal, we can conclude that the given function \(y=3t+t^2\) is indeed a solution to the differential equation \(ty'-y=t^2\).

Key concepts

Solutions to Differential Equations

Differential equations represent mathematical equations that relate a function with its derivatives. To solve a differential equation means finding a function or a set of functions that satisfy the equation.

The process of solving becomes more practical when applied to a real-world scenario, such as understanding the growth rate of a population or the cooling rate of a hot cup of coffee. For instance, with the differential equation provided, our task is to verify whether the given function, in this case,\( y = 3t + t^2 \), satisfies the equation:

• The differential equation is \( ty' - y = t^2 \).
• The given function is expressed as \( y = 3t + t^2 \).

The solution to this differential equation will be a function that, when plugged back into the equation, results in an identity – both sides of the equation being equal.

Knowing how to find solutions to differential equations allows us to model changes, predict future outcomes, or simply understand the underlying mechanics of dynamical systems.

Verification of Solutions

Verification is an essential step in validating whether a function truly satisfies a given differential equation. When dealing with differential equations, one cannot assume that a proposed solution is correct; it must be verified through substitution and algebraic manipulation.

The verification of our solution \( y = 3t + t^2 \) for the differential equation\( ty' - y = t^2 \) involves several steps:

• First, calculate the derivative of the proposed function. Here, \( y' = 3 + 2t \).
• Substitute both \( y \) and \( y' \) into the original differential equation.
• Simplify the expression \( t(3 + 2t) - (3t + t^2) \) to obtain \( t^2 \).

By confirming that both sides of the equation are equivalent, we verify that the function \( y = 3t + t^2 \) is indeed a solution to the differential equation.

This process ensures that the function satisfies the relationship described by the equation over the considered interval of \( t \). It also reinforces the understanding that a verified solution holds true, effectively modeling the behavior described by the differential equation.

First Order Differential Equations

First order differential equations involve derivatives that are first order, meaning they include the first derivative but no higher derivatives. These equations often take the general form:

• \( F(t, y, y') = 0 \) or \( y' = f(t, y) \).

In the exercise at hand, the differential equation is \( ty' - y = t^2 \). Here, we see that it establishes a relationship between the first derivative \( y' \) and the variables \( t \) and \( y \). A common characteristic of first-order differential equations is that they often model rates of change, like velocity or growth rates.

Solving first order differential equations typically involves finding an explicit function \( y \) in terms of \( t \) – a function that describes how quantities vary over time or another independent variable. The solution gives us insight into how a process develops and unfolds from given initial conditions or under certain constraints.

In practice, once you have a first-order equation, you may explore methods such as separation of variables, integrating factors, or even graphical solutions to find the solution function. Understanding these solutions can be crucial for problems in physics, engineering, and many fields where modeling real-world phenomena is required.