Please look in comments for cleaner version or ask me to post it again. dy/dt = k2 x(null)*e^(-k1*t)-k3*y/(y+M) I have no idea how to approach this diff EQ
this is implicit differentation, I can show you how to solve this but you need to write the equation in a way so it is easier to understand
Ok
just use equation editor use these brackets to show exponents {}
\[dy/dt= k_{2}\times x_{0} \times e^{-k_{1}*t} - (k_{3} y(t))\div(y+M)\]
Better?
yes :)
Ok this question looks more complicated than what I can handle you better ask in chat :L
Yeah, its the function of y that is throwing me off
Does anyone have any ideas of how to approach it? I'm stuck haha
dy/dt = k(2x^0)(e^-k1t) - (k3y/y) + M is this the equation?
Yeah sorry I guess the equation editor form doesn't work in the question field
\[dy/dt = k_{2} x_{0}e^{-k_{1}t}-k_{3}y/(y+M)\]
maybe this will help
wolfram happens to be terrible with many DE's
Yeah, I can't get it into any proper form
i know
and why did you write "differentiate" ?
Yeah, I don't think Wolfram alpha can do this one, I've already tried using DSolve as well
this is what my professor told me to do
just to write it super clear\[\large \frac{dy}{dt}=k_2x_0e^{-k_1t}-{k_3y\over y+M}\]and M, x0, and the k's are all constant?
Correct, how did you format the division like that?
there are two ways {x\over y}\or \frac{x}{y}
O, thanks. Any ideas for the problem?
If I had an idea I'd say it I'm just trying to make sense of it :/
O sorry, its an equation modeling alcohol consumption, y is a function of t and represents the level of alcohol in the blood at time t
Its a cascading problem where x(t) represents blood in the stomach/G.I. track, and I already solved for x(t) and plugged it into dy/dt
calling for help: @lalaly she's not online, you may have to post this the the section "L differential eqns", but that section is slow so you may have to wait a day or so for a response.
Alright, thanks a lot.
Sorry, how to I find that section?
Nvm found it
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