Question

Find the general solution. \(y+e^{t} y^{\prime}=0\)

Short answer

Answer: \(y(t) = Ce^{-e^t}\), where \(C\) is an arbitrary constant.

Step-by-step solution

Identify the ODE type

The given differential equation is: \(y+e^{t} y^{\prime}=0\) It is a first-order linear ordinary differential equation (ODE) of the form \(y' + a(t)y = 0\), where \(a(t) = e^t\).

Find the integrating factor

To find the integrating factor, we need to compute \(e^{\int a(t) dt}\). In this case, we have: \(e^{\int e^t dt} = e^{e^t + C}\) We can ignore the constant \(C\) as only the integrand is important for finding the integrating factor. So the integrating factor is: \(\mu(t) = e^{e^t}\)

Multiply both sides by the integrating factor

Now, we multiply both sides of the given ODE by the integrating factor: \(e^{e^t}y + e^{e^t}e^t y' = 0\)

Rewrite and integrate

Notice that the left side of the equation is the derivative of the product \((e^{e^t} y)\). We can rewrite the equation as: \(\frac{d}{dt}(e^{e^t}y) = 0\) Now, integrate both sides of the equation with respect to \(t\): \(\int \frac{d}{dt}(e^{e^t}y)dt = \int 0 dt\) This gives: \(e^{e^t}y = C\)

Find the general solution

To find the general solution, we can now solve for \(y\): \(y = Ce^{-e^t}\) The general solution for the given differential equation is: \(y(t) = Ce^{-e^t}\), where \(C\) is an arbitrary constant.

Key concepts

Integrating Factor Method

The integrating factor method is a powerful technique for solving first-order linear ordinary differential equations (ODEs). The method hinges on finding a special multiplier, the integrating factor, which once applied to both sides of the ODE, simplifies the process of integration. We look for an integrating factor of the form \( \mu(t) = e^{\int a(t) dt} \), where \( a(t) \) comes from the standard form of the ODE, \( y' + a(t)y = 0 \).

By applying this factor, the ODE is transformed in such a way that the left side becomes the derivative of a product. This clever manipulation converts the problem into a direct integration task, sidestepping the need to integrate a y-function directly, which may not always be feasible. The integrand function \( a(t) \) is the only component needed to determine the integrating factor, making constant terms such as \( C \) in \( e^{\int a(t) dt} = e^{e^t + C} \) superfluous for calculating \( \mu(t) \) itself.

First-order Linear ODE

A first-order linear ODE is an equation that contains a function \( y \) and its first derivative \( y' \), structured in a specific linear form. The general structure is \( y' + p(t)y = g(t) \), where \( p(t) \) and \( g(t) \) are functions of \( t \) alone. In such equations, \( y' \) is not raised to any power other than one and is not multiplied by \( y \) or any function of \( y \).

The given example \( y + e^{t} y' = 0 \) fits the description, with \( p(t) = e^t \) and \( g(t) = 0 \). These equations are prevalent because of their straightforward properties and the existence of systematic methods for solving them, such as the integrating factor method discussed previously.

General Solution of ODE

The general solution of an ordinary differential equation (ODE) encapsulates all possible solutions to the differential equation. It often includes an arbitrary constant (represented by \( C \)) because ODEs typically have infinite solutions that vary based on initial conditions. The final general solution \( y(t) = Ce^{-e^t} \) for the equation \( y + e^{t} y' = 0 \) embodies a family of functions, each corresponding to a different value of \( C \).

To determine the specific solution from a general solution, extra information—such as an initial value—is usually necessary. This course of action, identifying a particular solution that satisfies the initial condition from a general solution, is a fundamental part of solving ODE problems.

Exponential Function

The exponential function, commonly represented as \( e^x \) or \( exp(x) \) where \( e \) is Euler's number (approximately 2.71828), is significant in mathematics due to its unique properties. For instance, the rate of growth of the exponential function is directly proportional to its current value, making it a perfect tool for modeling phenomena with constant growth rates, such as compound interest in finance or certain natural processes in physics and biology.

In the context of differential equations, the exponential function often appears as part of the integrating factor or as a solution. Its behavior, with regard to differentiation and integration, is noteworthy: the derivative of \( e^x \) is \( e^x \) itself, and the integral of \( e^x \) is also \( e^x \) plus an arbitrary constant, making calculations involving exponentials much simpler and more tractable.