Does anyone know if the diff equation from pic related is solveable?

ha ha ha xd
no its non linear and looks cancerous as fuck. maybe try numerical methods?

File: 1.jpg (51KB, 800x541px) Image search: [Google]

51KB, 800x541px

I just plugged it into my ti89, it doesn't directly solve it, but it does reduce it to the form int(bunch of y shit dy)=x+C, if you care to see it. potato quality image

File: 2.jpg (52KB, 800x541px) Image search: [Google]

52KB, 800x541px

part 2, also used pi as the constant, and the @15 or @14 or whatever is the calculator displaying the constant of integration

and finally, heres a translation (sorry i don't know latex)

integral(1/[sqrt(2ln[y-pi]+2ln[y]+C)]dy) = x+C

>>8491211
you can integrate it once after multiplying through by y'. you get 1/2 (y')^2 on the left and log( y*(y-c) ) on the right. You could multiply through by 2 and take the square root to get y' on the left, but finding an anti-derivative of sqrt(log) seems pretty tough. Worth a shot maybe.

>>8491906
>>8491239
ha. I'm only as smart as a TI calculator.

File: 2016-11-23_16-29-49.gif (408KB, 1250x934px) Image search: [Google]

408KB, 1250x934px

>>8491906
>1/2 (y')^2 on the left and log( y*(y-c) ) on the right
how do you get that? lf l multiply by y' l get:

y''·y' = y'/y + y'/(y-c)

>integrate

y' = ln(y) + ln(y-c) + C

Pic related movement (x(t),y(t)) would describe being x(t) : x'' = 1/y + 1/(x-c) and y(t) y'' = 1/y + 1/(y-c). Kind of reminds me of the lissajous figures

>>8491929
nvm l'm retarded, it's 1/2 (y')^2

>>8491211
If you solve it numerically, it oscillates around (c/2). For y(0) close to c/2, it's close to sinusoidal; as y(0) gets larger, it becomes more triangular.