Question

Find the general solution to each differential equation. $$x y^{\prime}-e^{x}+y+x y=0$$

Short answer

\(y = \frac{e^x}{x} + \frac{C}{x}\), where \(C\) is the constant of integration.

Step-by-step solution

Write the differential equation in standard form

To find the general solution, first rewrite the given differential equation in standard form. The standard form for a first-order linear differential equation is: \(y' + P(x)y = Q(x)\). Put the given equation, \(xy' - e^x + y + xy = 0\), into this form by dividing through by \(x\) (assuming \(x eq 0\)): \[y' + \frac{1}{x}y = \frac{e^x}{x}\].

Identify the integrating factor

The next step is to find the integrating factor \(\mu(x)\) which is given by \(\mu(x) = e^{\int P(x)dx}\). Here, \(P(x) = \frac{1}{x}\), so we compute \(\int \frac{1}{x}dx = \ln|x|\). Thus, the integrating factor is \(\mu(x) = e^{\ln|x|} = |x|\). However, since the differential equation is only valid for \(x eq 0\), we can consider the integrating factor for \(x > 0\) to be \(x\), ignoring the absolute value.

Multiply through by the integrating factor

Now multiply every term of the differential equation by the integrating factor to get a new equation: \[x(y' + \frac{1}{x}y) = e^x\].

Recognize the left-hand side as a derivative product

The left-hand side of the equation from Step 3 can be recognized as the derivative of the product of the integrating factor and the unknown function, \((xy)'\). This allows us to write the equation as: \[(xy)' = e^x\].

Integrate both sides

Integrating both sides with respect to \(x\) gives \[\int (xy)'dx = \int e^x dx\], which simplifies to \[xy = e^x + C\], where \(C\) is the constant of integration.

Solve for \(y\)

Finally, to find \(y\) as a function of \(x\), simply divide both sides by \(x\) to get the general solution: \[y = \frac{e^x}{x} + \frac{C}{x}\].

Key concepts

First-Order Linear Differential Equation

Understanding a first-order linear differential equation is pivotal in learning differential equations. Such an equation relates a function, its first derivative, and the independent variable through a linear combination. Generally, they take the form
\(y' + P(x)y = Q(x)\).
To solve for the general solution, identifying this equation's structure is the essential first step. In the exercise, we initially have \(x y' - e^{x} + y + x y = 0\), which must be rearranged to fit the standard form by isolating terms involving the dependent variable \(y\) and its derivative \(y'\). Dividing through by \(x\) and reordering gives us \(y' + \frac{1}{x}y = \frac{e^x}{x}\). This conversion allows us to apply methods tailored for first-order linear equations to find the solution.

Integrating Factor

Moving on from recognizing the standard form, the concept of an integrating factor \(\mu(x)\) is vital in finding the solution to a first-order linear differential equation. The integrating factor is a mathematical device that, when multiplied by each term in the differential equation, transforms the left side into an exact derivative. It's usually of the form
\(\mu(x) = e^{\int P(x)dx}\),
where \(P(x)\) is the coefficient of \(y\) in the standard form of the differential equation. For the exercise in question, we calculate the integrating factor from \(P(x) = \frac{1}{x}\) by integrating \(\int P(x)dx\), resulting in \(\mu(x) = e^{\ln|x|} = |x|\). Since we're excluding \(x = 0\) due to division by \(x\), we simplify this to \(\mu(x) = x\) for \(x > 0\). This is a crucial step because it paves the way for achieving a differential equation that can be directly integrated to find the solution.

Separation of Variables

While the current exercise focuses on integrating factors, another fundamental approach for solving differential equations is the method of separation of variables. This technique is employed when a differential equation can be written as a product of functions solely in terms of \(x\) and \(y\), respectively. In simpler terms, it rearranges the equation so that one side has only \(x\)'s and differentials \(dx\), and the other side has only \(y\)'s and differentials \(dy\). Although this method is not directly applicable to our exercise, it's essential to recognize situations where separation of variables is the right tool. It's particularly useful in equations where the variables can be divided cleanly and integrated separately, leading to a general solution where constants of integration are determined by initial conditions or boundary values.