Second-order differential equation and conditions

In summary, the student is struggling with the last part of a question and is unsure if they have correctly interpreted the instructions. They provide the relevant equations and their attempt at a solution. Another user helps clarify the instructions and the student is able to understand the problem.
  • #1
Taylor_1989
402
14

Homework Statement


Hy guys I am have a problem with the last part of this question. part d), ii) I get the general formal which I have displayed below, but what I done understand is if I take the limits as show in ii) I get ##0=\
\infty## which obviously I am doing something wrong. Have I misinterpreted what they are say as y tends to 0 as x tends. Could anyone give me some advice it would be much appreciated, thanks in advance.
upload_2017-3-30_16-35-20.png

Homework Equations

The Attempt at a Solution


##y=c_1e^{2x}+c_2e^{-3x}-e^{-x}##
 

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  • #2
Taylor_1989 said:

Homework Statement


Hy guys I am have a problem with the last part of this question. part d), ii) I get the general formal which I have displayed below, but what I done understand is if I take the limits as show in ii) I get ##0=\
\infty## which obviously I am doing something wrong. Have I misinterpreted what they are say as y tends to 0 as x tends. Could anyone give me some advice it would be much appreciated, thanks in advance.
View attachment 115277

Homework Equations

The Attempt at a Solution


##y=c_1e^{2x}+c_2e^{-3x}-e^{-x}##
What equation did you get for question d) part i)? Part ii) is essentially saying that ##\lim_{x \to \infty} c_1e^{2x} + c_2e^{-3x} - e^{-x} = 0##. What are the values of the two constants so that this can happen?
 
  • #3
@Mark44 I see now thank you as soo as u put in that form and do each lim seperatly I saw what was going on. Much apprectied.
 

FAQ: Second-order differential equation and conditions

What is a second-order differential equation?

A second-order differential equation is a mathematical equation that involves the second derivative of a dependent variable with respect to an independent variable. It is commonly used to model physical systems and phenomena in fields such as physics, engineering, and economics.

What are the conditions for a second-order differential equation to have a unique solution?

In order for a second-order differential equation to have a unique solution, it must satisfy two conditions: 1) the equation must be linear, and 2) the initial values (or boundary values) of the dependent variable and its first derivative must be specified.

How are second-order differential equations solved?

There are several methods for solving second-order differential equations, including the following: 1) analytical methods, such as separation of variables and substitution, 2) numerical methods, such as Euler's method and Runge-Kutta methods, and 3) using software programs such as MATLAB or Maple.

What are some real-life applications of second-order differential equations?

Second-order differential equations are used to model a wide range of physical systems and phenomena, including the motion of objects under the influence of forces, electrical circuits, and population dynamics. They are also used in engineering to design and analyze structures, control systems, and other systems.

What is the difference between homogeneous and non-homogeneous second-order differential equations?

A homogeneous second-order differential equation has a zero constant term, while a non-homogeneous second-order differential equation has a non-zero constant term. In other words, a homogeneous equation can be written as a linear combination of the dependent variable and its derivatives, while a non-homogeneous equation cannot.

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