Problem 5 - Arbeitsbuch Mathematik für Ingenieure

Question

Lösen Sie die Anfangswertprobleme a) \(y^{\prime \prime}=e^{y}, \quad y(0)=0, y^{\prime}(0)=\sqrt{2}\), b) \(y^{3} \cdot y^{\prime \prime}+1=0, \quad y(1)=y^{\prime}(1)=1\).

Step-by-Step Solution

Answer

a) y = 2 \ln (1+\frac{x}{\sqrt{2}}) b) y = \sqrt{2x-1}

Step 1: Identify the given differential equation and initial conditions for part (a)

The given differential equation is \[y'' = e^{y}\]. The initial conditions are \[y(0) = 0\] and \[y'(0) = \sqrt{2}\].

Step 2: Solve the differential equation for part (a)

To solve \[y'' = e^{y}\], use the substitution \[v = y', v' = y''\]. Thus, the equation becomes \[v' = e^{y}\]. By separating variables,\[v dv = e^{y}dy\]. Integrate both sides to get \[\int v dv = \int e^{y} dy\]. The integration results in \[\frac{v^2}{2} = e^{y} + C_1\]. Using the initial condition \[v(0) = \sqrt{2}\], we get \[\frac{2}{2} = e^{0} + C_1 \Rightarrow 1 = 1 + C_1 \Rightarrow C_1 = 0\]. Therefore, \[\frac{v^2}{2} = e^y\]. Solving for \[v\], it follows that \[y' = \pm \sqrt{2e^y}\].

Step 3: Use initial conditions to find the particular solution for part (a)

From \[y'(0) = \sqrt{2}\], consider the positive root \[y' = \sqrt{2e^y}\]. Separating variables, \[\frac{dy}{\sqrt{2e^y}} = dx\]. Solving this equation, integrate both sides to obtain \[\int \frac{dy}{\sqrt{2e^y}} = \int dx \Rightarrow \frac{1}{\sqrt{2}}\int e^{-y/2}dy = x + C_2\]. Using the initial condition \[y(0) = 0\], it follows that \[0 = C_2\]. Therefore, the solution to the problem is \[y = 2 \ln (1+\frac{x}{\sqrt{2}})\].

Step 4: Identify the given differential equation and initial conditions for part (b)

The given differential equation is \[y^3 y'' + 1 = 0\]. The initial conditions are \[y(1) = 1\] and \[y'(1) = 1\].

Step 5: Solve the differential equation for part (b)

Rewrite as \[y^3 y'' + 1 = 0\Rightarrow y'' = -\frac{1}{y^3}\]. Using \[v = y'\Rightarrow v' = y''\], the equation becomes \[v \frac{dv}{dy} = - \frac{1}{y^3}\]. Separate variables to get \[v dv = -\frac{1}{y^3}dy\]. Integrate both sides: \[ \frac{v^2}{2} = \frac{1}{2y^2} + C_3\]. Using \[y'(1) = 1\], \[ \frac{1}{2} = \frac{1}{2(1)^2} + C_3 \Rightarrow 0 = C_3\]. Therefore, \[ \frac{v^2}{2} = \frac{1}{2y^2} \Rightarrow y' = \pm \frac{1}{y}\].

Step 6: Use initial conditions to find the particular solution for part (b)

\[y' = \frac{1}{y}\]. Using the method of separation of variables, \[y dy = dx\]. Integrate both sides \[ \frac{y^2}{2} = x + C_4\]. Using \[y(1)=1\], \[ \frac{1}{2}=1+C_4\Rightarrow C_4=-\frac{1}{2}\]. The final equation is \[ \frac{y^2}{2} = x - \frac{1}{2}\Rightarrow y^2 = 2x -1\]. Rearrange to obtain the particular solution \[y = \sqrt{2x-1}\].

Key Concepts

Understanding Differential Equations

Differential equations are equations involving the derivatives of a function. They describe how a quantity changes with respect to another. For instance, the second-order differential equation \[y'' = e^{y}\] indicates how the acceleration (second derivative) of a function relates to the exponential of the function itself.
Such equations are fundamental in modeling real-world phenomena, from physics to biology.
Solving these equations often involves integrating functions and utilizing initial conditions to find specific solutions.

Initial Value Problem (IVP)

An initial value problem (IVP) adds specific conditions to the differential equation. These conditions are usually given at the start (initial values).
For example, in the exercise, we saw initial conditions like \[y(0) = 0\] and \[y'(0) = \sqrt{2}.\] These conditions help in finding the unique solution to the differential equation that fits the given initial scenario.
Initial conditions play a crucial role in determining the behavior and trajectory of solutions in differential equations.

Integration Techniques

Integration is the process of finding the antiderivative of a function. It's used extensively for solving differential equations.
For instance, separating variables and integrating both sides is a common technique to solve differential equations:

• Separate variables: \[v dv = e^{y}dy\]
• Integrate both sides: \[\frac{v^2}{2} = e^y + C_1\]

By using the constants determined from initial conditions, we refine the general integral form into a particular solution.
Mastering integration techniques is essential for solving many types of differential equations.

Substitution in Differential Equations

Substitution is a powerful method used to simplify differential equations for easier solving.
In the given exercise, we used substitution to change variables:

• Let \[v = y' \Rightarrow v' = y''\]

This helped reformulate the equation into a simpler form:

• Substitute and simplify: \[v' = e^{y}\] becomes \[v dv = e^{y}dy\]

This reduces a complex equation into something more manageable, enabling easier integration and solution finding.
Knowing when and how to use substitution can be a game-changer in solving differential equations.

Separation of Variables

Separation of variables is a common method for addressing specific types of differential equations.
It involves isolating each variable on different sides of the equation to facilitate integration.

• For example, \[y'' = e^{y}\] was transformed using substitution to separate and integrate as \[v dv = e^{y}dy\]
• The next example \[y' = \frac{1}{y} \Rightarrow \ y dy = dx \]

Integrating both sides individually yields the function.
This technique is simple yet powerful, especially for solving first-order differential equations, making it a fundamental tool in the mathematician's toolkit.