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Chapter 3: Problem 6

Find the general solutions of the diferential equations in Problems. $$ y^{\prime \prime}+5 y^{\prime}+5 y=0 $$

Short Answer

Expert verified
The solution is \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \) with distinct roots \( r_1 = \frac{-5 + \\sqrt{5}}{2} \) and \( r_2 = \frac{-5 - \\sqrt{5}}{2} \).

Step by step solution

01

Identify Differential Equation Type

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \( y'' + 5y' + 5y = 0 \). Our goal is to find its general solution.
02

Write the Characteristic Equation

Replace \( y \) with \( e^{rt} \) (assuming solutions of the form \( y = e^{rt} \)) and convert the differential equation into a characteristic equation. This results in \( r^2 + 5r + 5 = 0 \).
03

Solve the Characteristic Equation

Solve the characteristic equation \( r^2 + 5r + 5 = 0 \) using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 5 \), and \( c = 5 \). Compute \( b^2 - 4ac = 5^2 - 4 \times 1 \times 5 = 25 - 20 = 5 \). Thus, the roots are \( r = \frac{-5 \pm \sqrt{5}}{2} \).
04

Write the General Solution

Since the roots \( r = \frac{-5 \pm \sqrt{5}}{2} \) are real and distinct, the general solution of the differential equation is \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \), where \( r_1 = \frac{-5 + \sqrt{5}}{2} \) and \( r_2 = \frac{-5 - \sqrt{5}}{2} \).

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Key Concepts

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

Characteristic Equation
A characteristic equation is essential in solving linear homogeneous differential equations with constant coefficients. It provides a way to transition from a differential equation to an algebraic one. This allows us to leverage simpler algebraic methods to find solutions.
  • Step 1: Identify the form of the differential equation, typically second-order or more complex. In our example, we have: \[ y'' + 5y' + 5y = 0 \]
  • Step 2: Assume a solution of the form \( y = e^{rt} \). This substitution simplifies things because the derivatives of \( e^{rt} \) relate directly to its original form, consistently involving \( r \), making differentiation straightforward.
  • Step 3: Substitute the assumed solution into the differential equation, transforming it into: \[ r^2 + 5r + 5 = 0 \]which is the characteristic equation.
The characteristic equation is a polynomial equation, usually quadratic, whose roots will determine the form of the solution to the original differential equation. It's a critical step that allows the conversion of a more complex calculus problem into manageable algebra.
Linear Homogeneous Equations
Linear homogeneous differential equations are a special category of equations that have several helpful properties.
  • The term 'linear' refers to how the terms in the equation are composed. A linear differential equation involves terms up to the first power only of the dependent variable and its derivatives. In our example:\[ y'' + 5y' + 5y = 0 \]
  • 'Homogeneous' indicates that the equation is equal to zero. This implies that if \( y(t) \) is a solution, then \( C \times y(t) \), where \( C \) is any constant, is also a solution. This superposition principle makes finding general solutions easier.
These types of equations are common in various fields, including physics and engineering, because they can effectively model systems subject to proportional resistance and restoring forces.Understanding that these equations describe how systems naturally tend to homogenize helps in conceptualizing the behavior of the solutions.
Quadratic Formula
The quadratic formula is a mathematical tool used to solve quadratic equations. A quadratic equation is typically written in the form \( ax^2 + bx + c = 0 \). Solving quadratics is crucial in finding the roots of the characteristic equation of a differential equation, as seen in our exercise.
  • To solve the characteristic equation, \( r^2 + 5r + 5 = 0 \), we apply the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = 5, \) and \( c = 5 \).
  • Substitute these values into the formula to compute the roots. In our problem:\[ b^2 - 4ac = 5^2 - 4 \times 1 \times 5 = 25 - 20 = 5 \]
  • The roots are then: \[ r = \frac{-5 \pm \sqrt{5}}{2} \]
The quadratic formula provides a precise method for finding the roots of any quadratic equation, crucial for determining the form of the general solution in differential equations. This step transforms and simplifies the task, bridging the algebraic findings to the calculus-based problem.

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Most popular questions from this chapter

In Problems, one solution of the differential equation is given. Find the general solution. $$ 3 y^{(3)}-2 y^{\prime \prime}+12 y^{\prime}-8 y=0 ; y=e^{2 x / 3} $$

Prove directly that the functions $$ f_{1}(x) \equiv 1, \quad f_{2}(x)=x, \quad \text { and } \quad f_{3}(x)=x^{2} $$ are linearly independent on the whole real line. (Suggestion: Assume that \(c_{1}+c_{2} x+c_{3} x^{2}=0\). Differentiate this equation twice, and conclude from the equations you get that \(c_{1}=c_{2}=c_{3}=0 .\) )

Consider the eigenvalue problem $$ y^{\prime \prime}+\lambda y=0 ; \quad y^{\prime}(0)=0, \quad y(1)+y^{\prime}(1)=0 $$ All the eigenvalues are nonnegative, so write \(\lambda=\alpha^{2}\) where \(\alpha \geqq 0\). (a) Show that \(\lambda=0\) is not an eigenvalue. (b) Show that \(y=A \cos \alpha x+B \sin \alpha x\) satisfies the endpoint conditions if and only if \(B=0\) and \(\alpha\) is a positive root of the equation \(\tan z=1 / z\). These roots \(\left\\{\alpha_{n}\right\\}_{1}^{\infty}\) are the abscissas of the points of intersection of the curves \(y=\tan z\) and \(y=1 / z\), as indicated in Fig. \(3.8 .13\). Thus the eigenvalues and eigenfunctions of this problem are the numbers \(\left\\{\alpha_{n}^{2}\right\\}_{1}^{\infty}\) and the functions \(\left\\{\cos \alpha_{n} x\right\\}_{1}^{\infty}\), respectively.

Suppose that the three numbers \(r_{1}, r_{2}\), and \(r_{3}\) are distinct. Show that the three functions \(\exp \left(r_{1} x\right), \exp \left(r_{2} x\right)\). and \(\exp \left(r_{3} x\right)\) are linearly independent by showing that their Wronskian $$ W=\exp \left[\left(r_{1}+r_{2}+r_{3}\right) x\right] \text { - }\left|\begin{array}{ccc} 1 & 1 & 1 \\ r_{1} & r_{2} & r_{3} \\ r_{1}^{2} & r_{2}^{2} & r_{3}^{2} \end{array}\right| $$ is nonzero for all \(x\).

(a) A uniform cantilever beam is fixed at \(x=0\) and free at its other end, where \(x=L\). Show that its shape is given by $$ y(x)=\frac{w}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2}\right) $$ (b) Show that \(y^{\prime}(x)=0\) only at \(x=0\), and thus that it follows (why?) that the maximum deflection of the cantilever is \(y_{\max }=y(L)=w L^{4} /(8 E I)\).
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