Differential equations questions

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 boundry conditions. Since those are not specified we can not go any further with the solution.
 
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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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