Problem 18 $$z^{\prime \prime}-2 z^{\prime}... [FREE SOLUTION]

Chapter 4: Problem 18

$$z^{\prime \prime}-2 z^{\prime}-2 z=0 ; \quad z(0)=0, \quad z^{\prime}(0)=3$$

Short Answer

Expert verified

The solution to the differential equation $z'' - 2z' - 2z = 0$ with initial conditions $z(0) = 0$ and $z'(0) = 3$ is $z(x) = \frac{3}{2\sqrt{3}}e^{(1 + \sqrt{3})x} - \frac{3}{2\sqrt{3}}e^{(1 - \sqrt{3})x}$

Step by step solution

Solve the Characteristic Equation

First, we'll need to solve the characteristic equation associated with the equation, which is given by $m^2 - 2m - 2 = 0$. By using the quadratic formula, $m = \frac{2 卤 \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = 1 卤 \sqrt{3}$. So, we have two solutions: $m = 1 + \sqrt{3}$ and $m = 1 - \sqrt{3}$

Write Down the General Solution

The general solution of a homogeneous second order differential equation with real, distinct roots $m_1$ and $m_2$ is $z(x) = c_1e^{m_1x} + c_2e^{m_2x}$. So for our equation, the general solution is $z(x) = c_1e^{(1 + \sqrt{3})x} + c_2e^{(1 - \sqrt{3})x}$

Apply the Initial Conditions

We need to find values for $c_1$ and $c_2$ that satisfy the initial conditions $z(0) = 0$ and $z'(0) = 3$. For $z(0)$, we have $c_1 + c_2 = 0$. For $z'(0)$, its derivative is $z'(x) = c_1e^{(1 + \sqrt{3})x}(1 + \sqrt{3}) + c_2e^{(1 - \sqrt{3})x}(1 - \sqrt{3})$. Substituting $x = 0$, we have $c_1(1 + \sqrt{3}) - c_2(1 + \sqrt{3}) = 3$. Solving this system of equations, we find that $c_1 = \frac{3}{2\sqrt{3}}$ and $c_2 = -\frac{3}{2\sqrt{3}}$

Write Down the Particular Solution

Substituting the values of $c_1$ and $c_2$ into the general solution, we get the particular solution satisfying the initial conditions: $z(x) = \frac{3}{2\sqrt{3}}e^{(1 + \sqrt{3})x} - \frac{3}{2\sqrt{3}}e^{(1 - \sqrt{3})x}$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation

When dealing with second order differential equations, the characteristic equation plays a crucial role in finding the solution. For the given differential equation $z'' - 2z' - 2z = 0$, its characteristic equation is derived by replacing $z''$ with $m^2$, $z'$ with $m$, and $z$ with 1: resulting in $m^2 - 2m - 2 = 0$.

This equation is a quadratic equation in terms of $m$, which helps us find the roots. These roots determine the nature of the solution to the differential equation.

Real and distinct roots lead to solutions in terms of exponentials.
Repeated roots result in exponential solutions with a polynomial factor.
Complex roots provide oscillatory solutions involving sine and cosine.

Initial Conditions

Initial conditions are given values for the function and its derivative at a specific point, usually $x = 0$, that allow us to find the particular solution from the general solution. For the example $z(0) = 0$ and $z'(0) = 3$, these conditions guide us in determining the constants $c_1$ and $c_2$ in the general solution.

Initial conditions are crucial as they tailor the general solution to fit specific scenarios.
The system of equations formed by initial conditions and the general solution aids in isolating the values of the constants.

Applying the given conditions ensures the solution satisfies the specific needs of the situation.

General Solution

The general solution of a second order differential equation is formed by finding the distinct solutions that satisfy the characteristic equation. For an equation such as ours with real, distinct roots $m_1 = 1 + \sqrt{3}$ and $m_2 = 1 - \sqrt{3}$, the general solution is expressed as:

\[ z(x) = c_1 e^{m_1 x} + c_2 e^{m_2 x} \]

Such solutions incorporate arbitrary constants $c_1$ and $c_2$, which are later specified by applying initial conditions.

The presence of the exponential terms highlights the influence of the roots on the behavior of the solution.
The solution format will vary depending upon the nature of the roots鈥攔eal, repeated, or complex.

Quadratic Formula

The quadratic formula is a fundamental algebraic tool to solve quadratic equations of the form $ax^2 + bx + c = 0$. For our characteristic equation $m^2 - 2m - 2 = 0$, using the quadratic formula:

\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, $a = 1$, $b = -2$, $c = -2$. Substituting these values gives:

\[ m = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-2)}}{2 \times 1} = 1 \pm \sqrt{3} \]

The quadratic formula provides precise solutions for polynomial roots.
It showcases the relationship between coefficients and the nature of the roots.

The formula thus bridges the transition from a polynomial characteristic equation to the determination of the differential equation's solution.

91影视

$$z^{\prime \prime}-2 z^{\prime}-2 z=0 ; \quad z(0)=0, \quad z^{\prime}(0)=3$$

Short Answer

Step by step solution

Solve the Characteristic Equation

Write Down the General Solution

Apply the Initial Conditions

Write Down the Particular Solution

Key Concepts

Characteristic Equation

Initial Conditions

General Solution

Quadratic Formula

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