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Chapter 3: Problem 6

Find the general solutions of the differential equations in Problems. \(y^{\prime \prime}+5 y^{\prime}+5 y=0\)

Short Answer

Expert verified
The general solution is \( y(t) = c_1 e^{\frac{-5 + \sqrt{5}}{2} t} + c_2 e^{\frac{-5 - \sqrt{5}}{2} t} \).

Step by step solution

01

Assume a Solution Form

For the given homogeneous linear differential equation, we assume a solution of the form: \[ y = e^{rt} \]where \( r \) is a constant to be determined.
02

Find the Characteristic Equation

Substitute \( y = e^{rt} \) into the differential equation \[ y^{\prime \prime} + 5y^{\prime} + 5y = 0 \]This gives:\[ (r^2e^{rt}) + 5(re^{rt}) + 5(e^{rt}) = 0 \]Simplifying, we have:\[ e^{rt}(r^2 + 5r + 5) = 0 \]Since \( e^{rt} eq 0 \), the characteristic equation is:\[ r^2 + 5r + 5 = 0 \]
03

Solve the Characteristic Equation

Solve the quadratic equation \( r^2 + 5r + 5 = 0 \) using the quadratic formula:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = 5, c = 5 \).Plug in these values:\[ r = \frac{-5 \pm \sqrt{25 - 20}}{2} = \frac{-5 \pm \sqrt{5}}{2} \]The roots are: \[ r_1 = \frac{-5 + \sqrt{5}}{2} \] and \[ r_2 = \frac{-5 - \sqrt{5}}{2} \]
04

Write the General Solution

Since the characteristic equation has two distinct real roots \( r_1 \) and \( r_2 \), the general solution of the differential equation is:\[ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} \]Substitute in the calculated roots:\[ y(t) = c_1 e^{\frac{-5 + \sqrt{5}}{2} t} + c_2 e^{\frac{-5 - \sqrt{5}}{2} t} \]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
The characteristic equation is a fundamental concept when working with homogeneous linear differential equations. It emerges from assuming a particular solution for the differential equation and leads us to a quadratic equation that helps us identify possible solutions.

Here's a brief overview of how it connects to solving differential equations:
  • We start with the differential equation, in this case, a homogeneous linear one: \(y'' + 5y' + 5y = 0\).
  • Assume a solution of the form \(y = e^{rt}\) where \(r\) is an unknown constant.
  • This form, when substituted back into the equation, simplifies to a product involving \(e^{rt}\) and a polynomial: \(r^2 + 5r + 5\).
  • The exponential function \(e^{rt} eq 0\), meaning the equation \(r^2 + 5r + 5 = 0\) must be solved; this is called the characteristic equation.
It is crucial since its roots determine the form of the general solution of the original differential equation. Solving this equation helps us understand behavior in systems modeled by differential equations.
Quadratic Formula
The quadratic formula is a means of solving quadratic equations of the form \(ax^2 + bx + c = 0\). This formula is essential when handling characteristic equations, allowing us to find exact solutions:
  • The characteristic equation \(r^2 + 5r + 5 = 0\) is in standard quadratic form, with \(a = 1\), \(b = 5\), and \(c = 5\).
  • Using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
  • Plugging in the values gives \[r = \frac{-5 \pm \sqrt{5}}{2}\], resulting in real and distinct roots.
  • The discriminant \(b^2 - 4ac\) here is positive, indicating two distinct real roots.
These roots feed directly into determining the form of the general solution for the differential equation, such as \(y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}\). Understanding how to apply the quadratic formula is vital for moving from the characteristic equation to a solution.
Homogeneous Linear Differential Equation
A homogeneous linear differential equation is a type of equation where each term is dependent on the solution and its derivatives only. The general form can be expressed as \(a_n y^{(n)} + a_{n-1} y^{(n-1)} + ... + a_1 y' + a_0 y = 0\).

In this exercise:
  • The given equation is \(y'' + 5y' + 5y = 0\), which is second-order because of \(y''\).
  • The absence of a non-zero function on the right-hand side confirms its homogeneity.
  • Solutions are generally approached by finding the characteristic equation and solving for \(r\).
  • The solution format \(y = c_1 e^{r_1 t} + c_2 e^{r_2 t}\) depends on the roots of this polynomial.
For students, this is a gateway into more complex analyses of dynamic systems. Recognizing a homogeneous equation and its general method of solution is fundamental. This awareness empowers tackling more complicated scenarios one might encounter in advanced studies or real-world applications.

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Most popular questions from this chapter

Solve the initial value problems in Problems 31 through \(40 .\) $$ y^{(4)}-y=5 ; y(0)=y^{\prime}(0)=y^{\prime \prime}(0)=y^{(3)}(0)=0 $$

Find general solutions of the equations in Problems. First find a small integral root of the characteristic equation by inspection; then factor by division. $$ y^{(3)}+3 y^{\prime \prime}+4 y^{\prime}-8 y=0 $$

Find the steady periodic solution \(x_{\mathrm{sp}}(t)=C \cos (\omega t-\alpha)\) of the given equation \(m x^{\prime \prime}+\) \(c x^{\prime}+k x=F(t)\) with periodic forcing function \(F(t)\) of frequency \(\omega .\) Then graph \(x_{\mathrm{sp}}(t)\) together with (for comparison) the adjusted forcing function \(F_{1}(t)=F(t) / m \omega .\) $$ x^{\prime \prime}+3 x^{\prime}+5 x=-4 \cos 5 t $$

Find and plot both the steady periodic solution \(x_{\mathrm{sp}}(t)=C \cos (\omega t-\alpha)\) of the given differential equation and the actual solution \(x(t)=x_{\mathrm{sp}}(t)+\) \(x_{\mathrm{tr}}(t)\) that satisfies the given initial conditions. $$ x^{\prime \prime}+4 x^{\prime}+5 x=10 \cos 3 t ; x(0)=x^{\prime}(0)=0 $$

Show directly that the given functions are linearly dependent on the real line. That is, find a nontrivial linear combination of the given functions that vanishes identically. $$ f(x)=e^{x}, g(x)=\cosh x, h(x)=\sinh x $$
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