Question

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=y\left(4 t^{3}+1\right), y(0)=4$$

Short answer

Question: Determine if the given first-order differential equation is separable and find the solution of the initial value problem. The equation is: $$y'(t) = y(4t^3 + 1), y(0) = 4$$ Answer: The given differential equation is separable, and the solution to the initial value problem is: $$y(t) = 4e^{t^4 + t}$$

Step-by-step solution

Checking for Separability

The given differential equation is: $$y'(t) = y(4t^3 + 1)$$ We can rewrite it as: $$\frac{dy}{dt} = y(4t^3 + 1)$$ This is already in the form \(\frac{dy}{dt} = P(t)Q(y)\), with \(P(t) = 4t^3 + 1\) and \(Q(y) = y\). As both \(P(t)\) and \(Q(y)\) only depend on \(t\) and \(y\) respectively, the equation is separable. #Phase 2: Solving the Initial Value Problem#

Separating Variables and Integrating

To solve the separable equation: $$\frac{dy}{dt} = y(4t^3 + 1)$$ Separate the variables by dividing by \(y\) and multiplying by \(dt\): $$\frac{1}{y} dy = (4t^3 + 1) dt$$ Now integrate both sides of the equation: $$\int \frac{1}{y} dy = \int (4t^3 + 1) dt$$

Evaluating the Integrals

Evaluate the integrals on both sides: $$\ln |y| = 4\int t^3 dt + \int dt$$ $$\ln |y| = t^4 + t + C$$

Applying Initial Condition and Solving for \(y\)

We are given the initial condition \(y(0) = 4\). Apply this condition to the equation: $$\ln |4| = 0^4 + 0 + C$$ $$\ln 4 = C$$ Now plug the value of \(C\) back into the equation: $$\ln |y| = t^4 + t + \ln 4$$ To find \(y(t)\), we use the inverse natural logarithm (exponential function): $$y(t) = e^{t^4 + t + \ln 4}$$ Now, we can simplify the exponent using the property \(e^a e^b = e^{a+b}\): $$y(t) = e^{t^4 + t} \cdot e^{\ln 4}$$ Since \(e^{\ln 4} = 4\), the final solution is: $$y(t) = 4e^{t^4 + t}$$

Key concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It focuses on two main concepts: differentiation and integration. Differentiation deals with how things change instantaneously, while integration concerns itself with the accumulation of quantities. In the context of this problem, calculus helps us understand how a rate of change (like a differential equation) can be used to describe and solve for the evolution of a quantity over time. We often encounter calculus when we model natural phenomena, such as motion, growth, or in this case, solving differential equations that describe how a quantity changes based on time.

Initial Value Problem

An initial value problem is a type of differential equation that comes with additional information in the form of an initial condition. This condition specifies the value of the solution at a particular point. In our equation, the initial value is given by \( y(0) = 4 \). This means that at time \( t = 0 \), the value of \( y \) is 4.

• The initial condition is crucial because it allows us to determine the constant of integration that appears when solving differential equations.
• Without an initial condition, there could be infinitely many solutions, but the initial condition ensures a unique solution that fits the specific scenario described.

In this problem, solving the initial value problem means finding a function \( y(t) \) that satisfies both the differential equation and the initial condition \( y(0) = 4 \).

Integration

Integration is a fundamental operation in calculus used to find a function from its derivative, essentially reversing the process of differentiation. In separable differential equations, once we have separated the variables, we integrate each side of the resulting equation. For the given exercise, integrating both sides:

• On the left side, \( \int \frac{1}{y} \, dy \) is integrated and results in \( \ln |y| \).
• On the right side, \( \int (4t^3 + 1) \, dt \) becomes \( t^4 + t + C \) after integration.

These integrals allow us to solve for \( y \) in terms of \( t \), and through the power of algebra, along with the exponential function, we retrieve the equation of the problem, ensuring it adheres to the initial conditions given.

Differential Equations

A differential equation is an equation that involves the derivatives of a function. In this problem, the function \( y(t) \) describes how \( y \) changes with respect to \( t \).

• Separable differential equations are a specific type that can be rewritten to allow variables to be separated on different sides of the equation.
• By doing so, the differential equation can be solved using integration.

Understanding and solving differential equations allows us to predict the behavior of dynamic systems over time. In this exercise, the equation \( y'(t) = y(4t^3 + 1) \) describes how \( y \) evolves, and by solving it, we gain insight into how \( y \) behaves as \( t \) varies. The final solution, \( y(t) = 4e^{t^4 + t} \), provides a complete picture of this dynamic relationship, revealing how \( y \) grows exponentially over time due to its dependence on both time and its own current state.