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Find the general solution of the following equations. $$\frac{d y}{d t}=2 y+6$$

Short Answer

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Question: Find the general solution for the first-order linear differential equation $$\frac{dy}{dt} = 2y + 6$$. Answer: The general solution for the given first-order linear differential equation is $$y(t) = -3 + Ce^{2t}$$, where C is an arbitrary constant.

Step by step solution

01

Identify the differential equation

The given differential equation is: $$\frac{dy}{dt} = 2y + 6$$
02

Rewrite the equation in standard form

The standard form of a first-order linear differential equation is: $$\frac{dy}{dt} + P(t)y = Q(t)$$ In our case, the given differential equation can be rewritten as: $$\frac{dy}{dt} - (-2y) = 2y + 6$$ So, we have \(P(t) = -2\) and \(Q(t) = 6\).
03

Find the integrating factor

The integrating factor, denoted as \(I(t)\), is given by: $$I(t) = e^{\int P(t) dt}$$ In our case, $$I(t) = e^{\int -2 dt} = e^{-2t}$$
04

Multiply the equation by the integrating factor

Now, we will multiply the entire differential equation by the integrating factor \(I(t) = e^{-2t}\): $$e^{-2t}\frac{dy}{dt} - 2e^{-2t}y = 6e^{-2t}$$
05

Use the product rule to rewrite the left side

Observe that the left side of the equation is the derivative of the product of \(y(t)\) and \(I(t)\): $$\frac{d}{dt}\left(y(t)e^{-2t} \right) = 6e^{-2t}$$
06

Integrate both sides of the equation

Now we will integrate both sides of the equation with respect to \(t\): $$\int\frac{d}{dt}\left(y(t)e^{-2t} \right)dt = \int 6e^{-2t}dt$$ Thus, we have: $$y(t)e^{-2t} = -3e^{-2t} + C$$
07

Isolate y(t) for the general solution

Finally, we will isolate \(y(t)\) to find the general solution: $$y(t) = -3e^{-2t}e^{2t} + Ce^{2t}$$ So, the general solution is: $$y(t) = -3 + Ce^{2t}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

first-order linear differential equation
A first-order linear differential equation is an equation that involves the first derivative of an unknown function, such as \( \frac{dy}{dt} \), and the function itself, \( y \), in a linear manner. The equation can be expressed as: \[ \frac{dy}{dt} + P(t)y = Q(t) \] where \( P(t) \) and \( Q(t) \) are functions of \( t \). The equation can describe many natural phenomena such as population growth, radioactive decay, and heat transfer. In essence, these equations represent processes where the rate of change depends linearly on time and state. Understanding the form of these equations facilitates their solution, often involving techniques like finding an integrating factor, as seen in our example.
integrating factor
The integrating factor is a crucial tool for solving first-order linear differential equations. It's a function, denoted by \( I(t) \), used to simplify the differential equation, making it easier to solve. The integrating factor is calculated as: \[ I(t) = e^{\int P(t) \, dt} \] Multiplying through the differential equation by \( I(t) \) transforms it into a form where the left-hand side becomes the derivative of a product, specifically \( \frac{d}{dt}(y(t)I(t)) \). This transformation is what allows us to integrate both sides with respect to \( t \) and eventually solve for the unknown function \( y(t) \). For our exercise, the integrating factor was \( e^{-2t} \). It helped simplify the differential into a format that aligned with the product rule for differentiation.
general solution
The general solution of a first-order linear differential equation encompasses all possible solutions of the equation. It is found after integrating and simplifying the equation using the integrating factor. For the equation given: \[ \frac{dy}{dt} = 2y + 6 \] Using the integrating factor and integrating, we derived: \[ y(t) = -3 + Ce^{2t} \] where \( C \) is an arbitrary constant determined by initial conditions or particular solutions. This result describes not a single solution but a family of solutions. Each value of \( C \) represents a unique function \( y(t) \) that satisfies the differential equation. The range of solutions reflects the unique conditions that the system can start from or reach.
ordinary differential equations (ODEs)
Ordinary differential equations (ODEs) are a type of differential equation involving functions of a single independent variable and their derivatives. They are classified by their order, which is the highest derivative present in the equation. ODEs play a critical role in modeling various real-world systems. For instance, they appear in physics for motion equations, in biology for growth models, and in economics for dynamic systems. Solving an ODE involves finding a function or set of functions that satisfy the equation. Our exercise specifically dealt with a first-order ODE, wherein only the first derivative of \( y \) with respect to \( t \) was considered. Techniques like the use of integrating factors help simplify and solve these equations, providing insights into the behavior of dynamic systems.

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