Question

Bestimmen Sie die allgemeine Lösung der Differentialgleichungen (BERNOULLI- Typ) (a) \(x y^{\prime}+y=y^{2} \ln x, x>0\), (b) \(x y^{\prime}+x y^{2}=y, x \neq 0\).

Short answer

(a) \( y = \frac{x}{\frac{x^2}{2} \ln x - \frac{x^2}{4} + C} \); (b) \( y = \frac{1}{Cx} \).

Step-by-step solution

Identify the Type of the Equation

Both equations are Bernoulli differential equations. These are equations of the form \( y' + P(x)y = Q(x)y^n \). In our exercise, \( n = 2 \).

Transform Equation (a)

Given equation (a) is \( x y^{\prime}+y=y^{2} \ln x \). Rewrite it as \( y' + \frac{1}{x} y = y^2 \ln x \). This matches the form \( y' + P(x)y = Q(x)y^n \) with \( P(x) = \frac{1}{x} \), \( Q(x) = \ln x \), and \( n = 2 \).

Substitute and Transform (a)

Introduce a substitution \( z = y^{1-n} = y^{-1} \), therefore \( z' = -y^{-2} y' \). Plug into the transformed equation \( y' + P(x)y = Q(x)y^2 \) to get \( -z' = P(x)z - Q(x) \). Substitute \( z = \frac{1}{y}, z' = -\frac{y'}{y^2} \) into \( x(-\frac{y'}{y^2}) + \frac{y}{x} = 1 \ln x \) and solve for \( z \).

Solve for z in Equation (a)

Equation becomes \( -z' = \frac{1}{x}z - \ln x \). Rearrange to \( z' + \frac{1}{x} z = \ln x \). This is a linear first-order differential equation, solve using an integrating factor \( \mu(x) = e^{\int \frac{1}{x} dx} = x \).

Integrate Equation (a)

Multiply through by the integrating factor: \( xz' + z = x \ln x \). The left side is the derivative of \( xz \), so integrate both sides: \( \int d(xz) = \int x \ln x \, dx \).

Solve the Integral for (a)

Solve \( \int x \ln x \, dx \) using integration by parts with \( u = \ln x, dv = x \, dx \), \( du = \frac{1}{x} \, dx, v = \frac{x^2}{2} \). This gives \( \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \, \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \).

Back-Substitute for y in (a)

You have \( xz = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \), with \( z = \frac{1}{y} \), so \( x \cdot \frac{1}{y} = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \). Solve for \( y \): \( y = \frac{x}{\frac{x^2}{2} \ln x - \frac{x^2}{4} + C} \).

Transform Equation (b)

Equation (b) is \( x y^{\prime} + x y^{2} = y \). Rewrite it as \( y' + y^2 = \frac{y}{x} \). This matches \( y' + P(x)y = Q(x)y^n \) with \( P(x) = 0 \), \( Q(x) = \frac{1}{x} \), and \( n = 2 \).

Substitute and Transform (b)

Introduce a substitution \( z = y^{-1} \), so \( z' = -y^{-2} y' \). Transform equation to \( -z' = 0 - \frac{z}{x} \).

Solve for z in Equation (b)

Equation becomes \( z' = \frac{z}{x} \). Rewrite as \( \frac{dz}{z} = \frac{1}{x} dx \) and integrate both sides.

Integrate for Equation (b)

Integrate: \( \int \frac{dz}{z} = \int \frac{1}{x} dx \), giving \( \ln |z| = \ln |x| + C \). Exponentiate: \( z = Cx \).

Back-Substitute for y in (b)

Since \( z = \frac{1}{y} \), we have \( \frac{1}{y} = Cx \), so \( y = \frac{1}{Cx} \).

Combine General Solutions

The general solutions are: (a) \( y = \frac{x}{\frac{x^2}{2} \ln x - \frac{x^2}{4} + C} \) and (b) \( y = \frac{1}{Cx} \).

Key concepts

Linear First-Order Differential Equations

Linear first-order differential equations are a simple yet powerful type of differential equation. These equations are of the form \( y' + P(x)y = Q(x) \), where \( y' \) is the derivative of \( y \) with respect to \( x \), and \( P(x) \) and \( Q(x) \) are functions of \( x \).
The goal is to find the function \( y(x) \) that satisfies this equation. The solution process often involves using an integrating factor, which we'll discuss further. In the solution provided for equation (a), once transformed, it becomes \( z' + \frac{1}{x} z = \ln x \), which is recognized as a linear first-order differential equation.
Understanding this concept helps students see these problems in various contexts and form a foundation for solving more complex differential equations.

Integration by Parts

Integration by parts is a technique used to solve integrals, especially when dealing with the product of two functions. This method is derived from the product rule of differentiation and is given by the formula:
\[\int u \, dv = uv - \int v \, du\]
In the exercise, to solve \( \int x \ln x \, dx \), we let \( u = \ln x \) and \( dv = x \, dx \). Differentiating and integrating these gives \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).
Substituting into the integration by parts formula, we get: \[\frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C\]where \( C \) is an integration constant. This essential concept allows one to solve integrals that seem complicated at first glance.

Integrating Factor

The integrating factor is a critical tool in solving linear first-order differential equations. An integrating factor, usually denoted as \( \mu(x) \), is strategically chosen to simplify the original differential equation.
For the equation \( z' + P(x)z = Q(x) \), the integrating factor is calculated as \( \mu(x) = e^{\int P(x) \, dx} \). With this factor, multiply the entire differential equation to transform it into something that can easily be integrated directly.

• In step 4 of the solution for equation (a), the integrating factor was found to be \( \mu(x) = x \), leading to the equation \( xz' + z = x \ln x \).

This transformation is pivotal, as it converts the left side into the derivative of a product, simplifying integration. It’s a structured approach that provides a consistent method for achieving solutions.

Differential Equation Transformation

Differential equation transformation is an effective technique to simplify and solve complex differential equations, often changing them into a more manageable form. Transformation typically involves substituting variables to change non-linear equations into linear ones.
In solving Bernoulli differential equations, a common transformation involves substituting \( z = y^{1-n} \). For instance, in this exercise, a substitution is made \( z = y^{-1} \), aiming to transform the original problem into a linear first-order equation.

• By this transformation, the non-linear component in equation (a) is removed, simplifying the solution process.
• Equations with straightforward linear forms can be solved using techniques like integrating factors.

Understanding how and why we transform equations is crucial for students, as it provides insight into the underlying structure of differential equations and preparation for more advanced topics.