Solving y=e^{t-t^2/2}: Luck or Methodology?

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In summary, the conversation discusses different methods of solving the equation y'' + ty' + ty = 0, including substituting y=uv and y=ekx, as well as using a Ricatti equation and the substitution k=q\frac{p'}{p}. The solution involves the functions c_1e^{\frac{-x^2}{4}+x}, c_2e^{\frac{-x^2}{4}+x} \int ^x e^{\frac{-s^2}{2}+2s}ds, and the error function Laplace transforms. The conversation also touches on the importance of providing enough information when seeking help with a problem.
  • #1
Wxfsa
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I know [itex]y=e^{t-t^2/2} [/itex] is a solution but how do I reach it? I tried substituting y=uv, making u' terms disappear then I noticed that the resulting u''=(__)u could be solved by a gaussian but this seems like luck to me
 
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  • #2
Have you tried substituting y=ekx?
 
  • #3
AdityaDev said:
Have you tried substituting y=ekx?
[itex]e^{kx}[/itex] cannot be a solution but I decided to try [itex]e^{k(x)}[/itex] which gave me a ricatti equation and after I substituted [itex]k=q\frac{p'}{p}[/itex] I got the original equation

edit: this is the solution
[itex]c_1e^{\frac{-x^2}{4}+x}+c_2e^{\frac{-x^2}{4}+x} \int ^x e^{\frac{-s^2}{2}+2s}ds[/itex]
 
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  • #4
A little nitpicky, but y'' + ty' + ty is not an equation (differential or otherwise), so it doesn't make any sense to talk about a solution. Is the full equation y'' + ty' + ty = 0?
 
  • #5
Mark44 said:
A little nitpicky, but y'' + ty' + ty is not an equation (differential or otherwise), so it doesn't make any sense to talk about a solution. Is the full equation y'' + ty' + ty = 0?
yes
 
  • #6
Mark44 said:
A little nitpicky, but y'' + ty' + ty is not an equation (differential or otherwise), so it doesn't make any sense to talk about a solution. Is the full equation y'' + ty' + ty = 0?
More nitpicking: Is y supposed to be a function of t or a function of x with t as a parameter?
 
  • #7
Svein said:
More nitpicking: Is y supposed to be a function of t or a function of x with t as a parameter?

you don't have to answer the question if you don't want to
 
  • #8
I actually solved this qustion after much work, however I do not like m solution. It would be great if someone could link me somewhere that derives error function laplace transforms

6hujRto.jpg
 

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  • #9
Svein said:
More nitpicking: Is y supposed to be a function of t or a function of x with t as a parameter?

Wxfsa said:
you don't have to answer the question if you don't want to
Of course that's true, but if you're asking for help with a problem, it's best to provide as much information about the problem as you can.
 

1. What is the equation y=e^{t-t^2/2} used for?

The equation y=e^{t-t^2/2} is commonly used in statistics and mathematical modeling to describe the probability density function of a normal distribution. It is also known as the Gaussian function or the bell curve.

2. Is solving y=e^{t-t^2/2} a matter of luck or methodology?

Solving y=e^{t-t^2/2} requires both luck and methodology. While luck may play a role in some cases, having a solid understanding of mathematical concepts and problem-solving techniques is crucial in successfully solving this equation.

3. Can y=e^{t-t^2/2} be solved using different methods?

Yes, there are multiple methods for solving y=e^{t-t^2/2}, such as using the quadratic formula, completing the square, or using the substitution method. The most appropriate method may depend on the specific problem and the individual's preferred approach.

4. What are some real-world applications of y=e^{t-t^2/2}?

The equation y=e^{t-t^2/2} has various applications in fields such as physics, finance, and biology. It is commonly used to model natural phenomena that follow a normal distribution, such as the height of a population or the amount of rainfall in a region.

5. Are there any shortcuts or tricks for solving y=e^{t-t^2/2}?

While there may be some shortcuts or tricks for solving y=e^{t-t^2/2}, it is essential to have a thorough understanding of the underlying concepts and techniques involved. It is always best to approach the problem systematically and avoid relying solely on shortcuts.

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