Differential equations questions

[Fritz]

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?

[arildno]

"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.

[Integral]

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 (http://www.physicsforums.com/showthread.php?t=8997).

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

To completly 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.

[Fady.bc]

i need help with this problem and please explain the answer:

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

thanks

[HallsofIvy]

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.