Question

Explain your selections. $$ y^{\prime}=y^{2} t^{3} $$

Short answer

The differential equation is separable and solves to: \( y = \frac{1}{-\frac{t^4}{4} - C} \).

Step-by-step solution

Identify the Type of Equation

The given equation is a first-order differential equation. It matches the form \( y' = f(y, t) \), where \( f(y, t) = y^2 t^3 \). This suggests it might be either separable or a Bernoulli equation.

Check for Separability

A separable differential equation can be expressed as \( g(y) \, dy = h(t) \, dt \). Our equation is \( y' = y^2 t^3 \). Rewriting the left side gives \( \frac{dy}{dt} = y^2 t^3 \), which can be rearranged as \( \frac{1}{y^2} \, dy = t^3 \, dt \). Thus, it is separable.

Integrate Both Sides

Integrate \( \frac{1}{y^2} \, dy \) and \( t^3 \, dt \) separately. The integral of \( \frac{1}{y^2} \) with respect to \( y \) is \( -\frac{1}{y} \). The integral of \( t^3 \) with respect to \( t \) is \( \frac{t^4}{4} \). Therefore, we have \( -\frac{1}{y} = \frac{t^4}{4} + C \), where \( C \) is the integration constant.

Solve for y

Rearrange the equation \( -\frac{1}{y} = \frac{t^4}{4} + C \) to solve for \( y \):First, multiply through by \( -1 \) to get \( \frac{1}{y} = -\left( \frac{t^4}{4} + C \right) \).Then, solve for \( y \) by taking the reciprocal: \( y = \frac{1}{-(\frac{t^4}{4} + C)} = \frac{1}{-\frac{t^4}{4} - C} \).

Key concepts

Separable Differential Equations

Separable differential equations are a type of equation in which the variables can be separated on opposite sides of the equation. Essentially, these equations allow us to rewrite the differential equation so that all terms involving one variable are on one side, and all terms involving the other variable are on the opposite side. This separation allows us to integrate each side independently, which simplifies solving the differential equation.

In the case of the given equation, \( y' = y^2 t^3 \), we separated the equation by rearranging it as \( \frac{1}{y^2} \, dy = t^3 \, dt \). This means we have expressed the equation in terms of \( g(y) \, dy = h(t) \, dt \), which confirms that it is separable.

The main advantage of a separable differential equation is this ability to isolate and integrate each variable independently, serving as the foundation for solving the equation.

First-Order Differential Equations

First-order differential equations are equations that involve the first derivative of a function but do not contain any higher-order derivatives. These kinds of equations can usually be solved by employing various methods such as separation of variables, as in our exercise, integrating factors, or other techniques.

The standard form for a first-order differential equation is \( y' = f(y, t) \). In our example, we see that the equation \( y' = y^2 t^3 \) fits into this form, with \( f(y, t) = y^2 t^3 \). This indicates that we are dealing with a first-order differential equation.

These equations not only model many real-world systems where the rate of change depends on current state but also serve as a stepping stone to understanding more complex equations.

Integration of Functions

Integration is a fundamental technique in calculus used to find functions when their derivatives are known, often referred to as solving differential equations. For separable differential equations, once we have separated the variables, each side of the equation is often solved by integrating each part separately.

In this particular example, we needed to integrate \( \frac{1}{y^2} \, dy \) and \( t^3 \, dt \). These yielded \( -\frac{1}{y} \) from integrating \( \frac{1}{y^2} \) and \( \frac{t^4}{4} \) from integrating \( t^3 \).

The process involves using integration rules and recognizing that you require an integration constant, \( C \), to complete the solution. Once we have integrated both sides, the result is an implicitly defined function that can sometimes be solved explicitly for one of the variables, as we did by solving for \( y \).

• Integral formulas are crucial for determining these antiderivatives.
• The integration constant helps account for initial conditions if provided.

Integrating correctly allows us to derive the relationships between variables that describe how one changes with respect to the other.