Problem 39 - Calculus Early Transcendentals

Question

Determine whether the following equations are separable. If so, solve the given initial value problem. $$\frac{d y}{d x}=e^{x-y}, y(0)=\ln 3$$

Step-by-Step Solution

Answer

If so, solve the initial value problem with the condition \(y(0) = \ln 3\). Answer: Yes, the given differential equation is separable. The solution of the initial value problem is $$y = \ln (e^x + 2)$$.

Step 1: Determine if the equation is separable

The given equation is $$\frac{d y}{d x}=e^{x-y}$$. We can rewrite it as $$\frac{d y}{d x}=e^xe^{-y}$$. The equation is now in the form \(\frac{dy}{dx}=g(x)h(y)\), with \(g(x)=e^x\) and \(h(y)=e^{-y}\). So, the given equation is separable.

Step 2: Separate the variables

To separate the variables, divide both sides of the equation by \(e^{-y}\) and multiply both sides of the equation by \(dx\). We get: $$e^y dy = e^x dx$$

Step 3: Integrate both sides

Now we can integrate both sides of the equation: $$\int e^y dy = \int e^x dx$$ The integrals are straightforward: $$e^y = e^x + C_1$$

Step 4: Apply the initial condition

To solve for the constant of integration \(C_1\), we apply the initial condition \(y(0) = \ln 3\). When \(x=0\), \(y=\ln 3\). $$e^{\ln 3} = e^0 + C_1$$ Simplify the equation: $$3=1+C_1$$ Solving for \(C_1\), we find that \(C_1=2\).

Step 5: Find the solution for y(x)

Now we can write the solution as: $$e^y = e^x + 2$$ Taking the natural logarithm of both sides: $$y = \ln (e^x + 2)$$ This is the solution of the given initial value problem.

Key Concepts

Initial Value Problem

An initial value problem in the context of differential equations is a condition that determines the specific solution to a differential equation from a general family of solutions. This is typically given in the form of the value of the unknown function and, occasionally, its derivatives at a particular point. In the case of the exercise \( \frac{dy}{dx} = e^{x-y}, y(0) = \ln 3 \) the initial value problem dictates the value of the function \( y \) when \( x \) is zero.

By knowing the value of the function at a specific point, in this case, \( y(0) = \ln 3 \) we are able to determine the constant of integration that makes the solution unique. This step is crucial because without the initial value, there would be infinitely many solutions due to the constant of integration. The process involves substituting the initial values into the integrated form of the equation and solving for the constant, as done in Step 4 of the solution provided.

Integrating Factors

Although integrating factors were not used directly in this exercise, they are a crucial concept in solving certain types of differential equations, such as linear non-separable equations. An integrating factor is a function that we multiply by the original differential equation to make it integrable. The idea is to alter the equation into a form where the left-hand side becomes the derivative of a product of functions which can then be integrated easily.

To find an integrating factor, one must often rely on a specific formula, usually relying on the coefficients of the differential equation. In the given exercise, the separation of variables was sufficient to solve the equation so an integrating factor was not necessary. However, understanding the concept can be essential when dealing with more complex differential equations where variables cannot be separated as cleanly.

Natural Logarithm

The natural logarithm, denoted as \( \ln \) is the logarithm to the base \( e \), where \( e \approx 2.71828 \), is the famous mathematical constant known as Euler's number. It is a fundamental function in calculus that allows us to solve equations where the variable is in an exponent, as seen in Step 5 of the solution.

In the provided solution, we needed to 'undo' the exponential function to solve for \( y \). We did this by applying \( \ln \) to both sides of the equation. A key property of logarithms used in this step is that \( \ln(e^x) = x \) for any \( x \) which follows from the definition of the natural logarithm. This property lets us simplify the solution to the very elegant form \( y = \ln(e^x + 2) \). The natural logarithm's inverse relationship with the exponential function makes it an indispensable tool in differential equations and many areas of applied mathematics.