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Consider a third order linear and homogeneous differential equation. How many arbitrary constants will its general solution involve?

Short Answer

Expert verified
Answer: The general solution of a third order linear and homogeneous differential equation involves 3 arbitrary constants.

Step by step solution

01

Definition of a third order linear and homogeneous differential equation

A third order linear and homogeneous differential equation is a differential equation that can be written in the form: dy^3/dt^3 + a2(dy^2/dt^2) + a1(dy/dt) + a0*y(t) = 0 where a0, a1, and a2 are constants and y(t) is the function we are trying to find.
02

General solution of a linear and homogeneous differential equation

The general solution of a linear homogeneous differential equation involves adding the homogeneous solutions together, which are found by solving the characteristic equation of the differential equation.
03

Characteristic equation for a third order linear homogeneous differential equation

To find the characteristic equation, we replace each derivative term with a power of the variable 'r'. For our third order linear homogeneous differential equation, the characteristic equation will be: r^3 + a2*r^2 + a1*r + a0 = 0
04

Solving the characteristic equation

Solving this cubic equation will give us three roots, which can be real or complex. These roots are denoted as r1, r2, and r3.
05

Homogeneous solutions

Based on the roots obtained from the characteristic equation, the homogeneous solutions can be written as: y1(t) = C1 * e^(r1*t) y2(t) = C2 * e^(r2*t) y3(t) = C3 * e^(r3*t) where C1, C2, and C3 are arbitrary constants.
06

General solution

The general solution of a third order linear and homogeneous differential equation is obtained by summing up the homogeneous solutions: y(t) = y1(t) + y2(t) + y3(t) y(t) = C1 * e^(r1*t) + C2 * e^(r2*t) + C3 * e^(r3*t)
07

Number of arbitrary constants

From the general solution above, we can see that there are 3 arbitrary constants: C1, C2, and C3. So, the general solution of a third order linear and homogeneous differential equation involves 3 arbitrary constants.

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Key Concepts

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

Third Order Differential Equation
A third order differential equation involves the third derivative of a function. It's like taking the idea of speed and acceleration a step further into what might be considered a kind of 'jerk' - change in acceleration.
In mathematical terms, a third order equation usually looks like this:
  • \( \frac{d^3y}{dt^3} + a_2\frac{d^2y}{dt^2} + a_1\frac{dy}{dt} + a_0y = 0 \)
Here, the term with the third derivative, \( \frac{d^3y}{dt^3} \), is what makes it third order. The constants \( a_0, a_1, \) and \( a_2 \) help influence the behavior of solutions.
Solving these equations helps find out how systems like mechanical engines respond to complex inputs over time. In simpler terms, it helps predict the motion and behavior of complex systems.
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is a function of the dependent variable and its derivatives. Nothing just hangs around without a derivative.
In these types of equations, substituting zero for the dependent variable makes the equation equal zero, which is not the case for non-homogeneous equations. Here’s an example:
  • \( y^{(3)} + a_2y'' + a_1y' + a_0y = 0 \)
These kinds of equations are critical in understanding natural systems where the output depends only on internal factors, like springs in mechanics or circuits governed only by resistors, capacitors, and inductors without any external input.
Characteristic Equation
The characteristic equation is a crucial concept used to simplify the process of solving linear differential equations. By substituting a trial solution of the form \( y = e^{rt} \) into our differential equation, we transform the problem into a polynomial, which is much easier to handle.
For a third order differential equation like:
  • \( r^3 + a_2r^2 + a_1r + a_0 = 0 \)
The roots of this polynomial are key to finding the solution to the differential equation. These roots, which can be real or complex, represent the behavior of the system over time. Finding these roots lets us know the natural frequencies or response times of the system at hand.
General Solution
The general solution of a differential equation is the group of all possible solutions derived from it. For a third order equation, the solution involves three arbitrary constants, reflecting the third order nature of the equation.
Once we've solved the characteristic equation for its roots, each root \( r_i \) gives a part of the solution associated with it:
  • \( y_1(t) = C_1 e^{r_1t} \)
  • \( y_2(t) = C_2 e^{r_2t} \)
  • \( y_3(t) = C_3 e^{r_3t} \)
The complete general solution is then the sum of these solutions:
  • \( y(t) = C_1 e^{r_1t} + C_2 e^{r_2t} + C_3 e^{r_3t} \)
The constants \( C_1, C_2, \) and \( C_3 \) allow for shifting and scaling - adjusting the solution to meet particular initial or boundary conditions. They ensure the solution fits the specific scenario you are analyzing, hence bringing the differential equation's abstract pattern into the reality of a particular system.

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Most popular questions from this chapter

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated in a surrounding where the ambient temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The pipe has a wall thickness of \(3 \mathrm{~mm}\) and an inner diameter of \(25 \mathrm{~mm}\), and it has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.003 \mathrm{~K}^{-1}\) and \(T\) is in \(\mathrm{K}\). Determine the outer surface temperature of the pipe.

What kind of differential equations can be solved by direct integration?

Consider a long solid cylinder of radius \(r_{o}=4 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the cylinder uniformly at a rate of \(\dot{e}_{\text {gen }}=35 \mathrm{~W} / \mathrm{cm}^{3}\). The side surface of the cylinder is maintained at a constant temperature of \(T_{s}=80^{\circ} \mathrm{C}\). The variation of temperature in the cylinder is given by $$ T(r)=\frac{\dot{e}_{\text {gen }} r_{o}^{2}}{k}\left[1-\left[1-\left(\frac{r}{r_{o}}\right)^{2}\right]+T_{s}\right. $$ Based on this relation, determine \((a)\) if the heat conduction is steady or transient, \((b)\) if it is one-, two-, or three-dimensional, and \((c)\) the value of heat flux on the side surface of the cylinder at \(r=r_{o^{*}}\)

Consider a homogeneous spherical piece of radioactive material of radius \(r_{o}=0.04 \mathrm{~m}\) that is generating heat at a constant rate of \(\dot{e}_{\text {gen }}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The heat generated is dissipated to the environment steadily. The outer surface of the sphere is maintained at a uniform temperature of \(110^{\circ} \mathrm{C}\) and the thermal conductivity of the sphere is \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the sphere, \((b)\) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and \((c)\) determine the temperature at the center of the sphere.

Consider a chilled-water pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and thermal conductivity \(k\). Water flows in the pipe at a temperature \(T_{f}\) and the heat transfer coefficient at the inner surface is \(h\). If the pipe is well-insulated on the outer surface, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe and \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation.

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