Problem 8 - Arbeitsbuch Mathematik für Ingenieure

Question

Die Riccati-Differentialgleichung $$ y^{\prime}=\left(-\frac{4}{x}-\frac{2}{x^{2}}\right) \cdot y+\frac{1}{x^{2}} \cdot y^{2}+\left(4+\frac{4}{x}+\frac{1}{x^{2}}\right) $$ besitzt für \(x>0\) die partikuläre Lösung \(u(x)=4 x+1\). a) Bestimmen Sie mit Hilfe des Ansatzes aus Satz \(2.4\) die allgemeine Lösung \(y(x)\). b) Welche dieser Lösungen erfüllt die Anfangsbedingung \(y(1)=6\) ?

Step-by-Step Solution

Answer

Using substitution and solving the auxiliary linear equation, the general solution is found, then applying \(y(1)=6\) provides the specific solution.

Step 1 - Inserting the particular solution

Insert the particular solution, \(u(x) = 4x + 1\), into the general Riccati equation. This verifies if the given function is indeed the solution.

Step 2 - Convert to auxiliary linear equation

Convert the Riccati equation to an auxiliary linear differential equation using substitution \( y(x) = u(x) + \frac{1}{v(x)} \). Plug in \( y = 4x + 1 \) into the original equation.

Step 3 - Simplification

After substitution and simplification, we get an auxiliary linear differential equation in terms of \(v(x)\). Simplify to find the resulting first-order linear ODE.

Step 4 - Solve the linear ODE

Solve the first-order linear differential equation obtained in Step 3. Find \(v(x)\) as a function of \(x\).

Step 5 - General Solution

Use the function \(v(x)\) to write the general solution of the original Riccati equation. Combine \(y(x) = u(x) + \frac{1}{v(x)}\) to find the general solution.

Step 6 - Apply the initial condition

Substitute the initial condition \( y(1) = 6 \). Use this to determine any constant in the general solution.

Step 7 - Final Solution

Having applied the initial condition, write the specific solution that satisfies \( y(1) = 6 \).

Key Concepts

Particular Solution

In differential equations, particularly Riccati differential equations, a particular solution is a specific solution that satisfies the given differential equation. For the given Riccati equation:\[ y' = \- \frac{4}{x} - \frac{2}{x^2} \cdot y + \frac{1}{x^2} \cdot y^2 + 4 + \frac{4}{x} + \frac{1}{x^2} \]The particular solution provided is \( u(x) = 4x + 1 \). The particular solution represents a specific function of \( x \) that directly solves the Riccati equation. Verifying if \( u(x) = 4x + 1 \) is indeed a solution involves substituting \( y = 4x + 1 \) back into the original equation to check if both sides of the equation match.

Auxiliary Linear Differential Equation

To solve the Riccati differential equation, we transform it into an auxiliary linear differential equation. This transformation simplifies solving by converting the non-linear Riccati equation into a form we can manage more easily. We use the substitution:\[ y(x) = u(x) + \frac{1}{v(x)} \]Here, \( u(x) = 4x + 1 \) is the particular solution. Substituting \( y \) in the Riccati equation, we derive a new equation in terms of \( v(x) \). The substitution transforms our Riccati equation into a simpler first-order linear differential equation in \( v \), making it easier to solve.

Initial Condition

Initial conditions are additional information provided to find a specific solution from a general one. In this problem, the initial condition given is \( y(1) = 6 \). This means when \( x = 1 \), the solution \( y(x) \) should equal 6. After converting to the auxiliary linear differential equation and finding the general solution, we use this initial condition to determine the unknown constants. Substituting \( x = 1 \) and \( y = 6 \) into the general solution helps identify the value of any constants, thus providing a specific solution tailored to the initial condition.

First-Order Linear ODE

A first-order linear ordinary differential equation (ODE) is an equation of the form:\[ y' + p(x)y = q(x) \]These are straightforward to solve using techniques like integrating factors. After transforming the original Riccati equation into an auxiliary linear differential equation, we get a first-order linear ODE in terms of \( v(x) \). Solving this ensures that we can find \( v(x) \) explicitly, which then helps in writing the general solution for \( y(x) \). The general solution combines the particular solution and the solution to the first-order linear ODE, giving a comprehensive solution to the initial Riccati equation.