Question

Use the methods of this section to solve the following differential equations. Compare computer solutions and reconcile differences. $$x y^{\prime}+y=e^{x y} \quad \text { Hint: Let } u=x y$$

Short answer

The solution is \(y = -\frac{1}{x} \ln|-(x + C)|\). Compare with computer solutions to verify.

Step-by-step solution

Substitute u = xy

Given the hint, let \(u = xy\). Then, differentiate both sides with respect to \(x\), we get \(u' = x y' + y\).

Rewrite Differential Equation

Substitute \(u\) and \(u'\) into the original differential equation, \(x y' + y = e^{xy} \). It becomes \(u' = e^u\).

Solve u' = e^u

Rewrite the equation as \(\frac{du}{dx} = e^u\). Separate the variables: \(\frac{1}{e^u} du = dx\). Integrate both sides: \(\int e^{-u} du = \int dx\).

Integrate Both Sides

The integral of the left side is \(-e^{-u}\), and the integral of the right side is \(x + C\). Thus, we have \(-e^{-u} = x + C\).

Solve for u

Rearrange to solve for \(u\): \(e^{-u} = -(x + C)\). Taking the natural logarithm on both sides gives: \(u = -\ln|-(x + C)|\).

Back-Substitute xy for u

Substitute back \(u = xy\): \(xy = -\ln|-(x + C)|\).

Isolate y

Solve for \(y\): \(y = -\frac{1}{x} \ln|-(x + C)|\).

Compare with Computer Solutions

Use computational tools (like MATLAB, Wolfram Alpha) to solve the differential equation numerically or symbolically and compare. Typically, the solutions should match the derived solution.

Key concepts

separation of variables

Separation of variables is a technique used to solve differential equations. The idea is to manipulate the equation so that each variable appears on a different side of the equation.

This allows us to integrate each side independently. For example, given the equation \( \frac{du}{dx} = e^u \), we can separate the variables as follows:

\[ \frac{1}{e^u} du = dx \]
This makes it easier to solve the equation through integration.

differentiation

Differentiation is a core concept in calculus. It is the process of finding the rate at which one quantity changes with respect to another. In our problem, we differentiated \(u = xy\) with respect to \(x\). Here's how:

Given \( u = xy \), we apply the product rule to get \( u' = x y' + y \).

This new equation helps us rewrite the original differential equation and simplifies our work in solving it.

integration

Integration is the reverse process of differentiation. It's used to find quantities when their rates of change are known. In the solution process, we integrated both sides of the separated equation:
\[ \frac{1}{e^u} du = dx \rightarrow \bullet \text{ Integrate both sides} \]
This integration gives us:

\[ \text{Left Side}: \text{Integral of } \frac{1}{e^u} du = -e^{-u} \]
\[ \text{Right Side}: \text{Integral of } dx = x + C \]
Combining these results, we got \( -e^{-u} = x + C \). This relation is key to solving our differential equation.

exponential function

The exponential function, written as \( e^x \), plays a significant role in calculus and differential equations. In our problem, we had the equation \( u' = e^u \). The exponential function here influences how we separate variables and integrate.

When integrating \( e^{-u} \), recall that the anti-derivative of \( e^u \) is itself. So, for \( e^{-u} \), the integral is \( -e^{-u} \).

Exponential functions exhibit continuous growth (or decay) which makes them crucial in modeling real-world phenomena like population growth or radioactive decay.

natural logarithm

The natural logarithm, denoted \( \text{ln}(x) \), is the inverse of the exponential function. In our solution, we used the natural logarithm to manipulate our results:

We rearranged: \( e^{-u} = -(x + C) \). To isolate \(u\), we applied the natural logarithm:

\( u = - \text{ln}|-(x + C)| \).

The natural logarithm simplifies expressions involving exponential functions and solves equations where growth rates are unknown.