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Chapter 9: Problem 5

Find the general solution. $$y^{\prime \prime}+2 y^{\prime}+5 y^{\prime}=0$$

Short Answer

Expert verified
The general solution for the given ODE \(y'' + 2y' + 5y = 0\) is \(y(x) = e^{-x}(A\cos(2x) + B\sin(2x))\), where \(A\) and \(B\) are arbitrary constants.

Step by step solution

01

Formulate the characteristic equation

The first step is to convert the given ODE into its characteristic equation, which is obtained by replacing the derivatives with the corresponding powers of a variable, say, \(r\). Therefore, the characteristic equation for the given ODE is: \(r^2 + 2r + 5 = 0\)
02

Solve the characteristic equation

Now, solve the quadratic equation obtained in step 1: \(r^2 + 2r + 5 = 0\) We can use the quadratic formula to find the roots: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In this case, \(a = 1\), \(b = 2\), and \(c = 5\). Plug the values into the formula: \(r = \frac{-2 \pm \sqrt{2^2 - 4(1)(5)}}{2(1)}\) This simplifies to: \(r = -1 \pm 2i\) These roots are complex numbers, which means our general solution will be in the form of exponentially decaying cosine and sine functions.
03

Set up the general solution

Since we have obtained complex roots, the general solution will be of the form: \(y(x) = e^{-x}(A\cos(2x) + B\sin(2x))\) Where \(A\) and \(B\) are arbitrary constants determined based on any specific initial conditions or boundary conditions. So, the general solution for the given ODE is: \(y(x) = e^{-x}(A\cos(2x) + B\sin(2x))\)

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Key Concepts

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

Characteristic Equation
For any ordinary differential equation (ODE), a common method to find solutions is to convert it into its characteristic equation. This process involves replacing derivatives in the equation with powers of a variable, typically denoted as \(r\).
  • For the equation \(y'' + 2y' + 5y = 0\), we replace \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with \(1\).
  • This results in a characteristic equation of \(r^2 + 2r + 5 = 0\).
The characteristic equation is a polynomial equation that is easier to handle. Solving it gives the roots and is a crucial step in determining the form of the solution to the original ODE.
Complex Roots
When solving a characteristic equation, the roots can be real or complex. In cases such as this exercise, the roots turn out to be complex.
  • Using the quadratic formula, \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the roots for \(r^2 + 2r + 5 = 0\).
  • Substituting \(a = 1\), \(b = 2\), and \(c = 5\) into the formula gives us \(r = -1 \pm 2i\).
Complex roots usually appear in conjugate pairs, which are \(r = -1 + 2i\) and \(r = -1 - 2i\) here. These roots result in a different form of the solution involving sine and cosine to incorporate the imaginary parts.
General Solution
Once the roots of the characteristic equation are found, constructing the general solution for the ODE is possible. When the roots are complex, the general solution involves exponential, sine, and cosine functions.
  • The real part of the root, \(-1\), contributes to the exponential decay factor \(e^{-x}\).
  • The imaginary part, \(2i\), determines the angular frequency of the sine and cosine functions, \(\cos(2x)\) and \(\sin(2x)\).
Thus, the general solution for our ODE is given by:\[ y(x) = e^{-x}(A\cos(2x) + B\sin(2x)) \]Here, \(A\) and \(B\) are arbitrary constants, which can be specified further when initial or boundary conditions are provided.

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Most popular questions from this chapter

Find the general solution. $$y^{\prime \prime}+2 y^{\prime}-8 y=0$$

A thermometer is taken from a room where the temperature is \(72^{\circ} \mathrm{F}\) to the outside, where the temperature is \(32^{\circ} \mathrm{F}\). Outside for \(\frac{1}{2}\) minute, the thermometer reads \(50^{\circ} \mathrm{F}\). What will the thermometer read after it has been outside for I minute? How many minutes does the thermometer have to be outside for at to read \(35 \mathrm{F}\) ?

(a) The differential equation $$ \frac{d P}{d t} \quad(2 \cos 2 \pi t) P $$ models a population that undergoes periodic fluctuations. Assume that \(P(0)=1000\) and find \(P(t) .\) Use a graphing utility to draw the graph of \(P\). (b) The differential equation $$ \frac{d P}{d t}=(2 \cos 2 \pi t) P+2000 \cos 2 \pi t $$ models a population that undergoes periodic fluctuations as well as periodic migration. Continue to assume that \(P(0)=1000\) and find \(P(t)\) in this case. Use a graphing utility to draw the graph of \(P\) and estimate the maximum value of \(P\)

Find the particular solution determined by the initial condition. $$y^{\prime}-y=e^{2 x}, \quad y(1)=1$$

Find the general solution. $$y^{\prime \prime}-13 y^{\prime}+42 y=0$$
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