# Y'' + ty' + ty

1. May 1, 2015

### Wxfsa

I know $y=e^{t-t^2/2}$ 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

2. May 1, 2015

Have you tried substituting y=ekx?

3. May 1, 2015

### Wxfsa

$e^{kx}$ cannot be a solution but I decided to try $e^{k(x)}$ which gave me a ricatti equation and after I substituted $k=q\frac{p'}{p}$ I got the original equation

edit: this is the solution
$c_1e^{\frac{-x^2}{4}+x}+c_2e^{\frac{-x^2}{4}+x} \int ^x e^{\frac{-s^2}{2}+2s}ds$

Last edited: May 1, 2015
4. May 1, 2015

### Staff: Mentor

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. May 1, 2015

yes

6. May 2, 2015

### Svein

More nitpicking: Is y supposed to be a function of t or a function of x with t as a parameter?

7. May 2, 2015

### Wxfsa

you don't have to answer the question if you don't want to

8. May 2, 2015

### Wxfsa

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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9. May 2, 2015

### Staff: Mentor

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.