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Differential Eq. Initial Value Problem (separable/partial fractions/hyperbolic)

  • Thread starter noob^n
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  • #1
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Homework Statement



dx/dt = 2*(1-x^2) with x(0)=0

Obtain the analytic solution to the IVP. Then, obtain the terms in the Taylor series up to and including the t^5 term.

Note: should be possible to express x(t) in terms of hyperbolic functions.

The Attempt at a Solution



some of my steps;

dx/(1-x^2) = 2dt

*partial fractions + integrated*

0.5[ ln|-x+1| + ln|x+1| ] = 2t + C

-x^2 +1 = A*exp(4t) , A>=0

x(t)=sqrt(1-exp(4t))

im not sure if this is right at all, as its simply been a while since ive looked at DE's. If it is right, i cant seem to figure out what combination of hyperbolic functions it relates to! any help would be much appreciated!
 

Answers and Replies

  • #2
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5,196

Homework Statement



dx/dt = 2*(1-x^2) with x(0)=0

Obtain the analytic solution to the IVP. Then, obtain the terms in the Taylor series up to and including the t^5 term.

Note: should be possible to express x(t) in terms of hyperbolic functions.

The Attempt at a Solution



some of my steps;

dx/(1-x^2) = 2dt

*partial fractions + integrated*

0.5[ ln|-x+1| + ln|x+1| ] = 2t + C
The work to here is fine, but the line below doesn't take into account the absolute value. Using the initial condition, you should have |1 - x2| = e4t.

For what x is the expression 1 - x2 nonnegative? Negative?
-x^2 +1 = A*exp(4t) , A>=0

x(t)=sqrt(1-exp(4t))

im not sure if this is right at all, as its simply been a while since ive looked at DE's. If it is right, i cant seem to figure out what combination of hyperbolic functions it relates to! any help would be much appreciated!
It seems to me that there is another part of this problem: to assume a solution in the form of a Maclaurin series, and to find its coefficients up to the x5 term. You're assuming a solution of the form x(t) = a0 + a1t + ... + a5t5 + ... + antn + ...

Basically, what you want to do is to substitute this into dx/dt + 2x2 - 2 = 0, and equate coefficients.
 
  • #3
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i thought because i conditioned A to be greater than or equal to zero, the RHS would stay positive, and i could drop the absolutes on the LHS... is there something wrong with my logic there?

all i really need to find is a suitable x(t) before i start subbing things into a maclaurin series... the hint said i should be able to write x(t) in terms of hyperbolic functions, and because none are springing out at me im getting the feeling im wrong.
 
  • #4
33,516
5,196
i thought because i conditioned A to be greater than or equal to zero, the RHS would stay positive, and i could drop the absolutes on the LHS... is there something wrong with my logic there?
Yes, there is. Use your initial condition to find A. You can't just arbitrarily say that A needs to be positive. Assuming that you find that A is positive (I did), you can't just get rid of the absolute values. Without them, 1 - x^2 is negative for some x and positive for others.
all i really need to find is a suitable x(t) before i start subbing things into a maclaurin series... the hint said i should be able to write x(t) in terms of hyperbolic functions, and because none are springing out at me im getting the feeling im wrong.
 
  • #5
5
0
Yes, there is. Use your initial condition to find A. You can't just arbitrarily say that A needs to be positive. Assuming that you find that A is positive (I did), you can't just get rid of the absolute values. Without them, 1 - x^2 is negative for some x and positive for others.
But because x is the dependant variable, we don't have to worry about crazy x values that cause errors... as all values of t (-inf, inf) will make x behave nicely. If the initial values didn't agree with the restraints on A then id agree there's a problem, but because they do I'm still not hugely convinced I'm wrong.

looking at 1-x2 = Ae4t , A>=0

x(0)=0 implies A=1

therefore x(t)=sqrt(1-exp(4t))

I know I must be sounding like a stubborn idiot, I apologize. I've got class tomorrow so I can run it through with the lecturer and post the answer.
 

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