Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Solve the following differential equations. For each differential equation, find the general solution and then find a solution passing through the point \((0, \sqrt{2})\). (a) \(\frac{d y}{d t}=-2 y\) (b) \(\frac{d y}{d t}=-2 t\) (c) \(\frac{d y}{d t}=-2\)

Step-by-Step Solution

Answer

The general solution and the solutions passing through the point \( (0, \sqrt{2}) \) are: \n (a) General solution: \( y = Ce^{-2t} \), Specific solution: \( y=\sqrt{2}e^{-2t}\) \n (b) General solution: \( y = -t^2 + C \), Specific solution: \( y=-t^2 + \sqrt{2}\) \n (c) General solution: \( y = -2t + C \), Specific solution: \( y=-2t+ \sqrt{2}\)

Step 1: Solve the Separable Differential Equation

The first equation is a separable differential equation. You can separate the variables and integrate both sides separately. \n Write like this \(-\int \frac{1}{y} dy = \int 2 dt\). After integration, you get \( - \ln |y| = 2t + C_1 \) where \( C_1 \) is the constant of integration. Solve for \( y \) to get \( y = e^{-2t - C_1} = C e^{-2t} \) where \( C = e^{-C_1} \) is a new arbitrary constant.

Step 2: Solve the Linear Non-separable Differential Equation

The second equation is a linear non-separable type. You separate the variables as much as possible which gives: \(\int dy = -\int 2t dt\). Solution is \( y = -t^2 + C \), where \( C \) is the constant of integration.

Step 3: Solve the Differential Equation with Constant Coefficients

The third equation has constant coefficients. It requires only basic integration. Solution is \( y = -2t + C \), where \( C \) is the constant of integration.

Step 4: Find a Solution Passing Through the Given Point

You now find the constants \( C \) for each differential equation that make the function pass through the point \( (0, \sqrt{2}) \). \n For \( y = Ce^{-2t} \), we have \(\sqrt{2}=Ce^0\) so \(C=\sqrt{2}\). So the solution is \(y=\sqrt{2}e^{-2t}\). \n For \( y = -t^2 + C \), we have \( \sqrt{2} = -0 + C \), thus \( C = \sqrt{2} \). So the solution is \(y=-t^2 + \sqrt{2}\). \n For \( y = -2t + C \), we gain \( \sqrt{2} = -2*0 + C \), therefore \( C = \sqrt{2} \). So the solution is \(y=-2t+\sqrt{2}\).

Key Concepts

Separable Differential Equations

When dealing with separable differential equations, you're looking at equations where you can rearrange to have all terms involving one variable on one side of the equation and the other variable on the opposite side. This allows you to integrate both sides separately.
For example, in the equation \( \frac{dy}{dt} = -2y \), you can rearrange the terms to form \(-\int \frac{1}{y} \, dy = \int 2 \, dt\). By integrating this equation, you find that \(-\ln |y| = 2t + C_1\).
This step is key because integration will give you an equation involving an integration constant \(C_1\), which can later be adjusted to suit an initial condition like a point through which the solution should pass.
The final form for the equation is \(y = Ce^{-2t}\), with \(C\) being derived from the initial condition, showing how separate integration of terms results in a solvable expression.

Linear Differential Equations

Linear differential equations are characterized by the presence of terms where the variables and their derivatives appear with a maximum degree of one. These are particularly straightforward to address because their algebraic structure allows the use of straightforward integration techniques.
Consider the example \( \frac{dy}{dt} = -2t \), which is a first-order linear differential equation. You can integrate both sides to find the general solution. After separating the variables, you integrate to obtain \( y = -t^2 + C \) where \(C\) is the constant of integration.
Linear differential equations are crucial in many practical applications because they often describe systems that change at a rate proportional to their current state. In this context, finding the constant \(C\) using an initial condition helps tailor the general solution to a specific scenario.

Constant Coefficients

Differential equations with constant coefficients are relatively simple because the coefficients of each term remain unchanged. This simplicity allows one to solve them using basic algebraic techniques and integration.
Let's look at \( \frac{dy}{dt} = -2 \). By integrating both sides, you easily find the general solution \( y = -2t + C \). Here, "constant coefficients" refers to the fact that coefficients in front of the derivative terms do not depend on the variables. These equations often resemble direct proportionalities and lend themselves to straightforward calculations.
Such problems often come with given points through which the curve must pass, like \((0, \sqrt{2})\). You use these to determine \(C\), giving a precise solution. They play a significant role in understanding systems with steady-state behavior or constant rates of change.