Question

Find the general solution to each differential equation. Try some by calculator. $$y^{\prime}=\frac{e^{x-y}}{e^{x}+1}$$

Short answer

The general solution to the differential equation \(y'=e^{2x}+y\) is \(y = e^{2x} + Ce^x\), where \(C\) is an arbitrary constant.

Step-by-step solution

Identify the Type of Differential Equation

Examine the given differential equation, which is a first-order linear ordinary differential equation in the form of \(y' - y = e^{2x}\).

Move All Terms Involving the Function y to the Left

Rewrite the differential equation to isolate terms that involve the function \(y\) on one side. The equation becomes \(y' - y = e^{2x}\).

Find an Integrating Factor

An integrating factor is typically \(e^{\int P(x)dx}\), where \(P(x)\) is the coefficient of \(y\) in the standard form. For the equation \(y'-y=e^{2x}\), \(P(x)=-1\). Therefore the integrating factor is \(e^{-x}\).

Multiply Every Term by the Integrating Factor

Multiplying both sides of the rewritten equation by \(e^{-x}\) yields \(e^{-x}y' - e^{-x}y = e^{x}\). This allows us to write the left side as the derivative of a product: \((e^{-x}y)'\).

Integrate Both Sides With Respect to x

Integrating both sides with respect to \(x\) gives us \(\int (e^{-x}y)' dx = \int e^x dx\), which simplifies to \(e^{-x}y = e^x + C\), where \(C\) is the constant of integration.

Solve for y

Finally, solve for \(y\) by multiplying both sides of the equation by \(e^x\) to get \(y = e^{2x} + Ce^x\), which is the general solution to the differential equation.

Key concepts

Integrating Factor

Understanding an integrating factor is crucial when dealing with first-order linear ordinary differential equations (ODEs). An integrating factor is a function that we multiply with the original ODE to turn it into an easily integrable form. This approach transforms a non-exact ODE into an exact one. In the context of our example, with the differential equation \(y' - y = e^{2x}\), we obtain the integrating factor by exponentiating the integral of the coefficient of \(y\), which is \(P(x) = -1\). The integrating factor thus becomes \(e^{-x}\). When we multiply each term of the original ODE by this integrating factor, we get an expression \(e^{-x}y' - e^{-x}y = e^{x}\) that allows us to proceed to integrate both sides with respect to \(x\).

Ordinary Differential Equation

An Ordinary Differential Equation (ODE) is an equation involving functions and their derivatives. In a first-order linear ODE, the equation involves only the first derivative of a function and the function itself, exemplified by \(y' = e^{2x} + y\). The objective is to find the function \(y\) that satisfies this relationship for all values of \(x\). ODEs are fundamental in expressing the relationships in dynamic systems, such as physics, engineering, and other applied sciences.

To solve the given ODE, we need to manipulate it into a form that enables us to apply the integrating factor, as already discussed. ODEs can reveal much about the phenomena they model, including general trends and specific behaviors as functions of the independent variables.

Constant of Integration

The constant of integration, denoted commonly as \(C\), is an essential component in the solution of differential equations. Whenever we integrate an indefinite integral, a constant of integration must be added to account for all the possible functions that could have produced the same derivative. In the step by step solution, after integrating \(\int (e^{-x}y)' dx = \int e^x dx\), we obtain \(e^{-x}y = e^x + C\), where \(C\) embodies all the potential constants that could have been lost in the derivative process. Significantly, \(C\) allows us to present the general solution, which includes all possible specific solutions to the differential equation.

Differential Equation Solution

The solution to a differential equation is the function or set of functions that satisfy the equation. In our example, after applying the integrating factor and integrating, we obtain \(e^{-x}y = e^x + C\). The last step in finding the differential equation solution is to solve for \(y\), which gives us the general solution \(y = e^{2x} + Ce^x\). This general solution encompasses all particular solutions that could arise from different initial conditions reflected in the value of \(C\). It's the culmination of the process we've followed: identifying the type of ODE, finding an integrating factor, applying it, integrating, and finally, isolating the dependent variable \(y\) to find its expression in terms of \(x\) and the constant \(C\).