| Kristian
D'Amato |
Is there any way to solve y for the following d.e? And if there is, could anyone please write how to do it. 2x - (dy/dx) = e^y. OR 2x - (1/y)(dy/dx) = y, which is much the same thing, The problem obviously lies in the e^y. If it where x instead of y it would have been easy. Thank you beforehand. I just require the answer to this d.e. for a little theory of mine. If it works out i will tell you all about it. I looked in several textbooks but couldnt find any similar one. I hope it is solvable. |
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| Arun
Iyer |
kristian, both of these equations are of the form of bernoulli's equation. love arun |
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| Arun
Iyer |
ok this time around i think i must be right!! For your first equation, i don't think there is a proper integrating factor to make the equation exact. For your second equation, since it is of the Bernoulli's form i get the solution as y ex2=òex2 dx + C however, the integral in between seems insoluble. love arun |
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| Marcos |
I think that both equations aren't expressible in closed form... I suppose we could construct a series though, as the solution... Is this what you want Kristian? Marcos |
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| Kristian
D'Amato |
What i wanted was a solution y = f(x) with no integration signs. I cannot have an infinite series, as that would mean accuracy only to a certain number of decimal places given that you only substitute values in the first few terms. Anyway, it seems that the equations are quite insoluble without using post-highschool methods. |
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| Marcos |
What was your idea though? I'm really interested to hear it (if you don't mind sharing it) Marcos P.S. You could always define a function to be the solution to this (and derive a few interesting properties it has) |
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| Andre
Rzym |
Kristian, I can see why you would prefer a solution in terms of standard functions, but practically, does it make much difference? If you think that when you compute sin(x) on a calculator, it is computing either the first few terms of an expansion, or using an approximation formula which is known to be accurate to a given precision. If you try a power series solution for y in the first equation, and try to match up coefficients, it will require a lot of effort and more importantly you will have difficulty in knowing what precision you have if you truncate to n terms. Instead, you could proceed as follows (which gives you a result similar to Arun's) 1) substitute y=x2-z to give dz/dx=ex2-z 2) rearrange: òez dz = òex2 dx 3) ez=C+ò0x ex2dx So y=x2-ln[C+ò0x ex2 dx] Now ex2 can be written as a power series, and so can the integral thereof. In particular, it is possible to put bounds on the error incurred in truncating to n terms. Andre |
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| Colin
Prue |
Is it safe to integrate term-by-term for something like ex^2 - is there ever a divergence issue? |
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| Andre
Rzym |
Good question. If the original series (in this case ex^2 ) converges for |x| < R, then the power series created by integrating term by term will converge similarly. In this particular case, the series converges for all finite x, and therefore so does the power series representing the integral. Andre |