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Solve the given differential equations. $$2 D^{2} y-3 D y-y=0$$

Short Answer

Expert verified
The general solution is \(y(x) = C_1 e^{\left(\frac{3 + \sqrt{17}}{4}\right)x} + C_2 e^{\left(\frac{3 - \sqrt{17}}{4}\right)x}\).

Step by step solution

01

Identify the Operators and Equation

Begin by identifying the differential operators in the equation. The given problem is \(2 D^{2} y - 3 D y - y = 0\). Here, \(D\) represents the derivative with respect to \(x\). The equation is a linear homogeneous differential equation with constant coefficients.
02

Convert to Characteristic Equation

To solve the differential equation, express it in terms of its characteristic equation. Replace each \(D^n\) with \(r^n\), where \(r\) denotes the roots of the characteristic equation. For the given equation, this becomes: \[2r^2 - 3r - 1 = 0\]
03

Solve the Characteristic Equation

Solve the characteristic equation for \(r\). The equation \(2r^2 - 3r - 1 = 0\) is a quadratic equation and can be solved using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a = 2\), \(b = -3\), and \(c = -1\).
04

Calculate the Roots

Calculate the roots using the quadratic formula:\[r = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\]\[r = \frac{3 \pm \sqrt{9 + 8}}{4}\]\[r = \frac{3 \pm \sqrt{17}}{4}\]Thus, the roots are \(r_1 = \frac{3 + \sqrt{17}}{4}\) and \(r_2 = \frac{3 - \sqrt{17}}{4}\).
05

Write the General Solution

The general solution of the differential equation is formed by using the roots of the characteristic equation. Since the roots are real and distinct, the general solution is:\[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]Substitute the values of \(r_1\) and \(r_2\):\[y(x) = C_1 e^{\left(\frac{3 + \sqrt{17}}{4}\right) x} + C_2 e^{\left(\frac{3 - \sqrt{17}}{4}\right) x}\]Where \(C_1\) and \(C_2\) are arbitrary constants.

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Key Concepts

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

Characteristic Equation
In differential equations, particularly when dealing with linear homogeneous differential equations with constant coefficients, the characteristic equation is a vital concept for finding solutions. Whenever we have such a differential equation, we can convert it into a characteristic equation. This is done by replacing the derivative operator, often denoted as \(D\), with a variable, typically \(r\).

This transformation changes the differential equation into a polynomial equation in terms of \(r\). For example, if the differential equation is \(2D^2 y - 3Dy - y = 0\), its characteristic equation becomes \(2r^2 - 3r - 1 = 0\). Once this polynomial equation is established, solving it will yield the roots that determine the form of the solution to the differential equation.

These roots, depending on their nature (real, distinct, repeated, or complex), guide us in forming the appropriate general solution.
Linear Homogeneous Differential Equation
A linear homogeneous differential equation with constant coefficients is a specific type of differential equation characterized by three main features:
  • Linearity: The equation is linear in terms of the dependent variable and its derivatives. This means that the equation does not contain terms like \(y^2, (Dy)^3\), etc.
  • Homogeneity: The equation is set equal to zero, indicating it is homogeneous. This means all terms involve the dependent variable or its derivatives.
  • Constant Coefficients: The coefficients of the differential terms (derivatives) are constants, simplifying the process of solving the equation.
These equations often take the form \(a_n D^n y + a_{n-1} D^{n-1} y + \, ... \, + a_1 Dy + a_0 y = 0\).

Solving these equations involves finding the characteristic equation and then determining its roots, which influence the structure of the general solution. These solutions typically involve exponential functions determined by the characteristic roots.
Quadratic Formula
The quadratic formula is key when tackling quadratic equations like the characteristic equation derived from a linear homogeneous differential equation. A quadratic equation takes the form \(ax^2 + bx + c = 0\). The quadratic formula, \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), gives us the solutions for the variable \(r\), which are the roots of the equation.

By substituting the coefficients \(a\), \(b\), and \(c\) from the characteristic equation into the quadratic formula, we can obtain the roots. These roots, once calculated, guide how we construct the particular solution to the differential equation.
  • Real and distinct roots result in exponential solutions multiplied by constants.
  • If roots are repeated, they lead to solutions that include a polynomial term multiplied by an exponential.
  • Complex roots provide solutions in terms of sine and cosine functions, reflecting oscillatory solutions.
Understanding and applying the quadratic formula accurately is crucial for solving characteristic equations effectively and providing the corresponding solutions for the differential equations in question.

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Most popular questions from this chapter

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