Question

Solving a First-Order Linear Differential Equation In Exercises \(51-58\) , solve the first-order differential equation by any appropriate method. $$ \left(2 y-e^{x}\right) d x+x d y=0 $$

Short answer

The solution of the given first-order linear differential equation is \( y= \frac{e^{x}(x-1) + C}{x^2}\).

Step-by-step solution

Writing the equation in differential form

Rearranging the terms to separate the differentials gives: \(dy = -\frac{(2y-e^x)}{x} dx \)

Rewriting Equation in Standard Form

The standard form of a first-order differential equation is \(dy/dx + P(x)y = Q(x)\). The given equation can be rewritten as: \( dy/dx + \frac{2}{x} y = \frac{e^{x}}{x} \). Here, \( P(x) = \frac{2}{x} \) and \( Q(x) = \frac{e^{x}}{x}\).

Find the Integrating Factor

The integrating factor is \( e^{\int P(x) dx} \). Therefore, the integrating factor is \( e^{\int 2/x dx} = x^2 \).

Multiply through by the integrating factor

Multiplying every term of the equation by \( x^2 \) gives: \( x^2 dy/dx + 2x y = x e^{x} \).

Find the solution

The left side is the derivative of \( yx^2 \) with respect to \( x \). So integrate the equation, yielding: \( \int (yx^2)' dx = \int x e^{x} dx \). Use integration by parts on the right hand side, where \( u=x \), \( dv=e^{x} dx \), \( du=dx \), and \( v=e^{x} \). So the integral of \(x \cdot e^{x} dx\) is equal to \(x \cdot e^{x} - e^{x}\). Therefore, the solution is \( yx^2= x e^{x} - e^{x} + C \). Solve for y to get the final solution: \( y= \frac{e^{x}(x-1) + C}{x^2}\).

Key concepts

Integrating Factor

When faced with a first-order linear differential equation, finding an integrating factor simplifies the equation and is key to solving it. An integrating factor is a function, often denoted as \( \mu(x) \), that makes the left side of the equation into the derivative of a product of functions.

In our example, the equation \( \frac{dy}{dx} + \frac{2}{x}y = \frac{e^x}{x} \) can be adapted using an integrating factor. We determine this factor by calculating \( e^{\int P(x) \, dx} \), where \( P(x) = \frac{2}{x} \).

• Compute the integral \( \int \frac{2}{x} \ dx = 2 \ln |x| \)
• Raise \( e \) to this power, resulting in the integrating factor \( x^2 \)

Using the integrating factor, multiply throughout the equation to transform it, substantiating the equation's left side into a perfect derivative.

Differential Form

Before employing any methods to solve differential equations, it's crucial to manipulate them into a differential form. This approach makes it easier to manage and apply further solution techniques. The differential form is commonly represented as \( M(x, y) \, dx + N(x, y) \, dy = 0 \).

In the original exercise, the provided equation was \((2y - e^x) \, dx + x \, dy = 0\).

• It denotes a relationship where components are expressed with respect to differentials \( dx \) and \( dy \).
• Rearranging terms enabled separating the differentials into a standard form \( dy = -\frac{(2y-e^x)}{x} \ dx \).

Integration by Parts

Integration by parts is a versatile technique for solving integrals. It can transform complex integrals into simpler ones by integrating the product of two functions. It is based on the formula: \[\int u \, dv = uv - \int v \, du\]

This technique was pivotal in the solution's final step. For \( \int x e^x \, dx \), we selected:

• \( u = x \), \( dv = e^x \, dx \)
• Then \( du = dx \), and \( v = e^x \)

Applying the formula gives us: \( x e^x - \int e^x \, dx = x e^x - e^x \). This calculation helped solve for the function \( y \).

Standard Form of Differential Equations

The standard form of a first-order linear differential equation is crucial for applying systematic solution methods. It is represented as \( \frac{dy}{dx} + P(x)y = Q(x) \). This organization aligns the equation for further operations, like locating an integrating factor or solving using specific methods.

In our case, the standard form derived was:
\[ \frac{dy}{dx} + \frac{2}{x} y = \frac{e^x}{x} \]

• This format highlights that \( P(x) = \frac{2}{x} \)
• And \( Q(x) = \frac{e^x}{x} \)

The transformation to this form is an essential pre-step in solving the differential equation and ensures consistency in applying methods like integrating factors.