Differential equations questions

In summary: Well, (x-1)(x+1)2= 0 has exactly those roots. (D-1)(D+1)2= 0 will have those roots for any polynomial in D. So (D-1)(D+1)2y= 0 is one linear d.e. with x and cosh x as solutions.
  • #1
Fritz
65
0
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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  • #2
"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.
 
  • #3
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 boundry conditions. Since those are not specified we can not go any further with the solution.
 
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  • #4
i need help with this problem and please explain the answer:

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

thanks
 
  • #5
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.
 

FAQ: Differential equations questions

What are differential equations?

Differential equations are mathematical equations that describe how a quantity changes over time. They involve a function and one or more of its derivatives, which represent the rate of change of the function.

What are the applications of differential equations?

Differential equations are used in many fields of science and engineering to model various physical phenomena such as motion, heat transfer, population dynamics, and electrical circuits. They are also used in economics and finance to make predictions and analyze data.

How do you solve a differential equation?

There is no single method for solving all types of differential equations. The method used depends on the type of equation and its properties. Some common techniques include separation of variables, substitution, and using differential equation solvers.

What is the order of a differential equation?

The order of a differential equation is the highest derivative present in the equation. For example, a first-order differential equation has only the first derivative, while a second-order differential equation has both the first and second derivatives.

How are differential equations classified?

Differential equations can be classified based on their order, linearity (whether the dependent variable and its derivatives appear linearly or not), and whether they are ordinary or partial (depending on the number of independent variables). They can also be categorized as initial value problems or boundary value problems.

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