Differential equations questions

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Discussion Overview

The discussion revolves around solving various differential equations (D.E.s), including first-order and linear D.E.s. Participants seek clarification on methods for deriving solutions and understanding the structure of general solutions, as well as specific examples of D.E.s with given solutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expresses confusion about solving the equation ln 1/[(y^2)-1] and seeks clarification on arriving at the answer 1/2 ln[(y-1)/(y+1)].
  • Another participant notes that the expression y = Ae^-4x + Be^-6x represents the general solution of a D.E. and contains two arbitrary constants.
  • A different participant suggests that to derive the D.E. from the general solution, one should compute the first and second derivatives and check if they satisfy the D.E.
  • One participant asks for help in finding a linear D.E. that has x and cosh x as solutions, indicating a need for a characteristic equation with specific roots.
  • Another participant reiterates the request for help with the same problem and emphasizes the importance of starting a new thread for new questions.
  • A participant explains that since cosh(x) can be expressed in terms of exponential functions, the problem can be reframed to finding a linear D.E. with x, e^x, and e^-x as solutions.

Areas of Agreement / Disagreement

Participants generally agree on the need for clarification on methods for solving D.E.s, but there is no consensus on specific solutions or methods to apply. Multiple competing views and approaches are presented without resolution.

Contextual Notes

Some participants note the necessity of initial or boundary conditions to fully specify the arbitrary constants in the general solution, which are not provided in the discussion.

Fritz
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I was trying to solve a 1st order D.E. and couldn't solve this:

ln 1/[(y^2)-1] I know the answer is 1/2 ln[(y-1)/(y+1)], but could figure out how this is so.

Also, I don't understand what method you'd use to get from y = Ae^-4x + Be^-6x to d2y/dx2 +10 dy/dx +24y + 0. I can eliminate the arbitrary constants in simpler D.E.'s, but not this one! Can someone go through a logical method?


Also, ytan(dy/dx) = (4 + y^2) sec^2(x). How, in this case, would you go about separating the variables?
 
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"Also, I don't understand what method you'd use to get from y = Ae^-4x + Be^-6x to d2y/dx2 +10 dy/dx +24y = 0. I can eliminate the arbitrary constants in simpler D.E.'s, but not this one! Can someone go through a logical method?"
Eeh, the expression for y(x) is called the GENERAL solution for the DE.
It SHALL have two arbitrary constants in its expression.
 
Zero is a perfectly good constant.

I have been looking at the first part of this problem but really do not understand the problem. Please give a concise statement of the DE. You may want to investigate our LaTex equation engine.

As for the second part simply compute the first and second derivative of the given solution, does it satisfy the DE?

To completely specify A and B you also need 2 initial or boundary conditions. Since those are not specified we can not go any further with the solution.
 
Last edited:
i need help with this problem and please explain the answer:

Find a linear DE that has x and cosh x as solutions.

thanks
 
Fady.bc said:
i need help with this problem and please explain the answer:

Find a linear DE that has x and cosh x as solutions.

thanks
In future do not "hijack" other peoples threads to ask new questions. Start your own thread.

Since cosh(x)= (ex+ e-x)/2, this is the same as asking for a linear d.e. with x, ex, and e-x as solutions. Since x= xe0, any linear d.e. with x as a solution must have characteristice equation with 0 as a double root.

You need a linear d.e. having a characteristic equation with 0 as a double root and 1 and -1 as roots.
 

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