i need to solve

by saying that

and Y(t)=y(x(t))

then i need to find from them expresions for \fracxdy/dx and which are expresions of dY/dt similar type

i tried like this:

Y(t)=y(x(t))

so

what now

?

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- December 2nd 2010, 07:47 AMtransgalacticsolving by substituting a complex function
i need to solve

by saying that

and Y(t)=y(x(t))

then i need to find from them expresions for \fracxdy/dx and which are expresions of dY/dt similar type

i tried like this:

Y(t)=y(x(t))

so

what now

? - December 2nd 2010, 08:14 AMAckbeet
I'm not sure you're doing the second derivative properly. I agree with the first derivative, but you have to use the chain rule on the second derivative as well. See here.

- December 2nd 2010, 08:21 AMtransgalactic
i replaces e^t with x

so when i do the chain rules in the second derivative dx(t)/dt=e^t=x

that how i got x^2

i cant see the mistake

in general i dont see how this method works - December 2nd 2010, 08:30 AMAckbeet
You've correctly got

and hence

Next, you compute

Thus,

and therefore

Substituting into your DE yields

When you simplify, lots of nice things happen. Does this make sense? - December 2nd 2010, 08:34 AMtransgalactic
my prof said that x=e^t

and when i see dx/dt i replace it with x

because dx/dt=e^t=x

so there isnt supposed to be e^t in your solution - December 2nd 2010, 08:40 AMAckbeet
Well, continue simplifying, and substitute back in so you just have x's. It's six of one, a half-dozen of the other.

Incidentally, this method of solution is inferior to the usual Cauchy-Euler method, because with this solution, you are artificially constraining x to be positive, whereas that is not the case with the usual Cauchy-Euler method. - December 2nd 2010, 08:46 AMtransgalactic

ok i did the substitution

how it became easier

? - December 2nd 2010, 08:48 AMAckbeet
That is incorrect. How did you get that from my post # 4?

- December 2nd 2010, 08:55 AMtransgalactic

4 members canceled each other

i cant see my mistake

and its still a second order equation with x's as coefficient

i cant see how it got easier - December 2nd 2010, 08:59 AMAckbeet
You made two mistakes. The first one was in not noticing that the multiplies

**both**terms in the expression The second mistake is the mysterious that multiplies the as your last term on the LHS. That seems to come out of thin air. The multiplying the term does NOT multiply the term.

Perhaps both of these mistakes are a result of not noticing the parentheses carefully enough? You absolutely**have**to pay attention to details like parentheses if you're going to avoid mistakes. - December 2nd 2010, 09:19 AMtransgalactic
ok here i go carefull on every step

first i open the parentheses and put x^2 instead of

but Y is a function of t

not x

how to go further - December 2nd 2010, 09:22 AMAckbeet
Correct. Now, here's where I would differ with your professor, perhaps. You should have the exponentials on the RHS, not x^2. Then you've got a second-order linear non-homogeneous ode with constant coefficients. Turn the crank.

- December 2nd 2010, 09:33 AMtransgalactic
ok but how this method differs the y=x^a substitution method

?

my prof said that the exponent method is better because it solves equations that

x^a cant

in wikipedea they dont say anything about the difference between them

Cauchy eiler link

on what type i need to use what method

? - December 2nd 2010, 09:40 AMAckbeet
I'm not overly familiar with the differences between the two methods, other than the observation I made before. If you have a bona-fide Cauchy-Euler equation, then the x^a method should solve it. The exponential substitution might work for other types of DE's, but as I said, you're restricting yourself to positive x's. That might not matter so much if your x is really a t, and you're only interested in positive time.

- December 8th 2010, 09:58 AMtransgalactic
cant understand the transition from

to