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

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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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Have you tried substituting y=ekx?
 
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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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?
 
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
 
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?
 
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
 
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

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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.