Problem 3
Question
Lösen Sie die Differentialgleichung $$ y^{\prime \prime}+2 y \cdot\left(y^{\prime}\right)^{3}=0 $$ durch Angabe der Umkehrfunktion \(x(y)\).
Step-by-Step Solution
Verified Answer
Substitute and solve to separate variables, integrate twice to find the implicit solution for \( x(y) \).
1Step 1 - Identify the nature of the differential equation
The given equation is a second-order non-linear differential equation: \( y'' + 2y(y')^3 = 0 \). To solve this, look for a way to simplify it.
2Step 2 - Introduce a substitution
Let \( y' = p \), which implies that \( y'' = p \frac{dp}{dy} \). Substitute these into the differential equation to get \( p \frac{dp}{dy} + 2y p^3 = 0 \).
3Step 3 - Separate variables
Rearrange and separate variables: \( p \frac{dp}{dy} = -2y p^3 \). Dividing both sides by \( p^3 \), we get \( \frac{1}{p^2} dp = -2y dy \).
4Step 4 - Integrate both sides
Integrating both sides with respect to their variables, \( \int \frac{1}{p^2} dp = \int -2y dy \). This gives \( -\frac{1}{p} = -y^2 + C \), where C is the constant of integration.
5Step 5 - Solve for \( p \)
Simplify the integrated equation to solve for \( p \): \( \frac{1}{p} = y^2 - C \). Thus, \( p = \frac{1}{y^2 - C} \).
6Step 6 - Substitute back \( y' \)
Recall that \( p = y' = \frac{dx}{dy} \). So, \( \frac{dx}{dy} = \frac{1}{y^2 - C} \).
7Step 7 - Integrate to find \( x(y) \)
Separate the variables and integrate again: \( dx = \frac{1}{y^2 - C} dy \). Integrate both sides to get \( x = \int \frac{1}{y^2 - C} dy \). This integral will give the implicit solution involving \( x \) and \( y \).
8Step 8 - Interpret the final expression
The solution \( x(y) \) will be the result of the integral and include a constant of integration. The final answer should be an implicit or explicit function of \( x \) in terms of \( y \).
Key Concepts
second-order differential equationnon-linear differential equationseparation of variablesintegration in differential equations
second-order differential equation
A second-order differential equation involves the second derivative of an unknown function. In simpler terms, it's an equation that includes the term \( y'' \) (the second derivative of \( y \) with respect to \( x \)).
For instance, the given differential equation is: \( y'' + 2y(y')^3 = 0 \). Here, \( y'' \) is the second derivative. Typically, such equations are more complex to solve compared to first-order differential equations.
Solving a second-order differential equation often involves additional steps like introducing substitutions or integrating multiple times to find the general or particular solution.
For instance, the given differential equation is: \( y'' + 2y(y')^3 = 0 \). Here, \( y'' \) is the second derivative. Typically, such equations are more complex to solve compared to first-order differential equations.
Solving a second-order differential equation often involves additional steps like introducing substitutions or integrating multiple times to find the general or particular solution.
non-linear differential equation
A non-linear differential equation includes terms where the unknown function and its derivatives appear to higher powers or are multiplied together.
In our exercise, the equation: \( y'' + 2y(y')^3 = 0 \) is an example of a non-linear differential equation.
Here's why: the term \( 2y(y')^3 \) involves the product of \( y \) and \( (y')^3 \), making it non-linear. Such equations are generally harder to solve than linear ones and often require specific techniques or methods for their solutions.
In our exercise, the equation: \( y'' + 2y(y')^3 = 0 \) is an example of a non-linear differential equation.
Here's why: the term \( 2y(y')^3 \) involves the product of \( y \) and \( (y')^3 \), making it non-linear. Such equations are generally harder to solve than linear ones and often require specific techniques or methods for their solutions.
separation of variables
Separation of variables is a method used to solve differential equations by isolating each variable on opposite sides of the equation. This simplification often turns a complex differential equation into basic integrals.
In the solution for our problem, we separate variables in Step 3: \( p \frac{dp}{dy} = -2y p^3 \) becomes \( \frac{1}{p^2} dp = -2y dy \).
By doing this, we make the equation easier to integrate, helping us move toward finding the solution.
In the solution for our problem, we separate variables in Step 3: \( p \frac{dp}{dy} = -2y p^3 \) becomes \( \frac{1}{p^2} dp = -2y dy \).
By doing this, we make the equation easier to integrate, helping us move toward finding the solution.
integration in differential equations
Integration is a crucial step in solving differential equations. It enables us to find the general forms of the unknown functions.
For example, in our step-by-step solution, we integrate \( \frac{1}{p^2} dp \text{ and } -2y dy \) to get \( -\frac{1}{p} = -y^2 + C \).
Later, we also integrate the equation \( dx = \frac{1}{y^2 - C} dy \) to find the relationship between \( x \text{ and } y \).
Integration helps us move from a rate of change (derivative) to the function itself, which is essential in solving differential equations.
For example, in our step-by-step solution, we integrate \( \frac{1}{p^2} dp \text{ and } -2y dy \) to get \( -\frac{1}{p} = -y^2 + C \).
Later, we also integrate the equation \( dx = \frac{1}{y^2 - C} dy \) to find the relationship between \( x \text{ and } y \).
Integration helps us move from a rate of change (derivative) to the function itself, which is essential in solving differential equations.
Other exercises in this chapter
Problem 1
a) Bestimmen Sie alle Lösungen der Differentialgleichung \(y^{\prime \prime}=\sqrt{x \cdot y^{\prime}}\). b) Bestimmen Sie alle Lösungen der Differentialgleichu
View solution Problem 2
a) Lösen Sie das Anfangswertproblem $$ y^{\prime \prime}+y^{\prime}=x+2 \quad, \quad y(0)=y^{\prime}(0)=2 $$ b) Berechnen Sie die allgemeine Lösung der Differen
View solution Problem 4
Berechnen Sie die Lösung des Anfangswertproblems $$ y^{\prime \prime}=(y+1) \cdot y^{\prime}, y(1)=1, y^{\prime}(1)=2 $$
View solution Problem 5
Lösen Sie die Anfangswertprobleme a) \(y^{\prime \prime}=e^{y}, \quad y(0)=0, y^{\prime}(0)=\sqrt{2}\), b) \(y^{3} \cdot y^{\prime \prime}+1=0, \quad y(1)=y^{\p
View solution