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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 long homogeneous resistance wire of radius \(r_{o}=\) \(0.6 \mathrm{~cm}\) and thermal conductivity \(k=15.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\). The heat generated is transferred to water at \(100^{\circ} \mathrm{C}\) by convection with an average heat transfer coefficient of \(h=3200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wire, \((b)\) obtain a relation for the variation of temperature in the wire by solving the differential equation, and \((c)\) determine the temperature at the centerline of the wire.

Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(180 \mathrm{~W} / \mathrm{cm}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(12.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(13.5 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(64.7 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(180 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(191 \mathrm{~W} / \mathrm{cm}^{2}\)

Consider uniform heat generation in a cylinder and a sphere of equal radius made of the same material in the same environment. Which geometry will have a higher temperature at its center? Why?

Consider a large plane wall of thickness \(L=0.8 \mathrm{ft}\) and thermal conductivity \(k=1.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). The wall is covered with a material that has an emissivity of \(\varepsilon=0.80\) and a solar absorptivity of \(\alpha=0.60\). The inner surface of the wall is maintained at \(T_{1}=520 \mathrm{R}\) at all times, while the outer surface is exposed to solar radiation that is incident at a rate of \(\dot{q}_{\text {solar }}=300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\). The outer surface is also losing heat by radiation to deep space at \(0 \mathrm{~K}\). Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached.

Exhaust gases from a manufacturing plant are being discharged through a 10 - \(\mathrm{m}\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\), wall thickness of \(10 \mathrm{~cm}\), and thermal conductivity of \(40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The exhaust gases are discharged at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure specific heat of the exhaust gasses is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular day, the outer surface of the exhaust stack experiences radiation with the surrounding at \(27^{\circ} \mathrm{C}\), and convection with the ambient air at \(27^{\circ} \mathrm{C}\) also, with an average convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are 0.9. Assuming steady one-dimensional heat transfer, (a) obtain the variation of temperature in the exhaust stack wall and (b) determine the inner surface temperature of the exhaust stack.

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