Problem 6 - Arbeitsbuch Mathematik für Ingenieure

Question

Gegeben sei die Bernoullische Differentialgleichung $$ y^{\prime}=\frac{1}{2 x} \cdot y+6 x^{2} \cdot y^{5} $$ a) Transformieren Sie diese Differentialgleichung in eine lineare Differentialgleichung (vgl. Satz 2.3). b) Bestimmen Sie die allgemeine Lösung der Differentialgleichung für \(x>0\)

Step-by-Step Solution

Answer

The transformed linear differential equation is \[ v^{\backprime} + \frac{2}{x}v = -24x^{2}. \] The general solution is \[ y(x) = \bigg( -6 x^{2} + \frac{C}{x^{2}} \bigg)^{-\frac{1}{4}}. \]

Step 1 - Identify the Bernoulli Differential Equation

The given differential equation is \[ y^{\backprime}=\frac{1}{2x} \backcdot y + 6x^{2} \backcdot y^{5}. \] Identify that it has the form of a Bernoulli differential equation: \[ y^{\backprime} + P(x)y = Q(x)y^{n}, \] where \( P(x) = -\frac{1}{2x} \), \( Q(x) = 6x^{2} \) and \( n = 5 \).

Step 2 - Substitute with v = y^{-4}

Substitute \( v = y^{-4} \). This implies \( y = v^{-\frac{1}{4}} \) and calculate \( y^{\backprime} \) using the chain rule: \[ y^{\backprime} = -\frac{1}{4} v^{-\frac{5}{4}} v^{\backprime}. \]

Step 3 - Transform the equation

Substitute \( y \) and \( y^{\backprime} \) back into the original equation. This gives us: \[ -\frac{1}{4} v^{-\frac{5}{4}} v^{\backprime} = \frac{1}{2x}v^{-\frac{1}{4}} + 6x^{2}v^{-\frac{1}{4}}. \] Simplify to get: \[ v^{\backprime} + \frac{2}{x}v = -24x^{2}. \]

Step 4 - Solve the linear differential equation for v

This is now a linear differential equation in \( v \). To solve, find the integrating factor \( \backmu(x) \) which is: \[ \backmu(x) = e^{\backint \frac{2}{x} dx} = e^{2\backln x} = x^{2}. \] Multiply through by the integrating factor: \[ x^{2} v^{\backprime} + 2x v = -24 x^{4}. \] Integrate both sides with respect to \( x \): \[ x^{2} v = -6 x^{4} + C \rightarrow v = -6 x^{2} + \frac{C}{x^{2}}. \]

Step 5 - Transform back to y

Recall \( v = y^{-4} \), so: \[ y^{-4} = -6x^{2} + \frac{C}{x^{2}} \rightarrow y = \bigg( -6 x^{2} + \frac{C}{x^{2}} \bigg)^{-\frac{1}{4}}. \] Hence, the general solution is: \[ y(x) = \bigg( -6 x^{2} + \frac{C}{x^{2}} \bigg)^{-\frac{1}{4}}. \]

Key Concepts

Linear Differential Equations

Let's start with understanding what a linear differential equation is. A differential equation is called linear if it can be written in the form \[ a(x)y' + b(x)y = c(x) \] where \( y \) is the dependent variable, \( x \) is the independent variable, and \( a(x) \), \( b(x) \), and \( c(x) \) are functions of \( x \). Importantly, the dependent variable \( y \) and its derivatives are only to the first power and are not multiplied together. In the exercise, we first transformed a Bernoulli equation into this linear form, which is key to making it more manageable and solvable with standard techniques. Linear differential equations have a predictable form that we can use integrating factors to solve.

Integrating Factor

The integrating factor is a powerful method to solve linear differential equations. For an equation in the form \[ y' + P(x)y = Q(x), \] the integrating factor \( \backmu(x) \) is given by \( e^{\backint P(x) \backspace dx} \). This factor is designed to simplify the equation by making the left-hand side a derivative of a single term. In our exercise, we found \( \backmu(x) = x^{2} \) for the transformed linear equation. Multiplying through by \( x^{2} \) created a straightforward path to integration and simplified our work.

Chain Rule

The chain rule is fundamental in calculus for differentiating composite functions. When we perform substitutions, like \( v = y^{-4} \) as in the exercise, we need the chain rule to correctly express the derivatives. The chain rule states that if \( y = f(g(x)) \), then \( y' = f'(g(x)) \backcdot g'(x) \). In the exercise, we had to take the derivative of \( v = y^{-4} \), resulting in \[ y' = -\frac{1}{4} v^{-\frac{5}{4}} v^{\backprime}. \] This correct application of the chain rule helped us transform the original Bernoulli differential equation.

General Solution

Finding the general solution of a differential equation means solving for the dependent variable in terms of the independent variable, incorporating an arbitrary constant. After transforming and solving the linear differential equation, we obtained \( v(x) = -6 x^{2} + \frac{C}{x^{2}} \). By reversing the substitution \( v = y^{-4} \), we found\[ y = \bigg( -6 x^{2} + \frac{C}{x^{2}} \bigg)^{-\frac{1}{4}}. \] This function represents the general solution which includes any particular solutions depending on the constant \( C \).

Transformation of Equations

Transformations simplify differential equations to a form that is easier to handle. For Bernoulli differential equations, we use the substitution \( v = y^{1-n} \) to convert it to a linear differential equation. In this exercise, the transformation \( v = y^{-4} \) turned a complex non-linear equation into something linear and tractable. Such transformations are crucial methods because they bridge our way from complex equations to forms where we can apply well-known solving techniques like integrating factors.