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alyannahere
Group Title
hi ! can you help me solve this ?
Find the general or particular solution for this differential equation
1. dy/dt=2y
2. dy/dx = 0.5y; y(0) = 100
i realized that it's all y so i don't really know how to go about that.
 one year ago
 one year ago
alyannahere Group Title
hi ! can you help me solve this ? Find the general or particular solution for this differential equation 1. dy/dt=2y 2. dy/dx = 0.5y; y(0) = 100 i realized that it's all y so i don't really know how to go about that.
 one year ago
 one year ago

This Question is Closed

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
First one :) \[\huge \frac{dy}{dt}=2y\] Let's multiply both sides by dt, and divide both sides by 2y
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
\[\huge \frac{dy}{2y}=dt\] Now integrate both sides :)
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
@alyannahere ?
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
solving it... :)
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
not really sure how.. ?
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
i'm really sorry.. i don't know how to go about this problem @PeterPan
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Well, maybe if it's tweaked a bit, you'll find it simpler... \[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\] Sorry about the other one, it's a typo o.O
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
so that is 1/2 1/y = C ?
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
No... remember \[\huge \int\limits \frac{dx}{x}=\ln(x) + C\]
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
so it's 1/2 ln y = C ?
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
my friend said the answer is y=C e ^2t
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
i don't know how he got it
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Hang on, just integrate for now, and tell me what you got :)\[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\]
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
1/2 ln y = C ?
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Well, isn't it though \[\huge \int\limits dx = x + C\]
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
how did it become dx ? i'm so sorry i just really don't understand this lesson
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
This is just a reminder :) The integral of 1 is just x (plus a constant)
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
then ?
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Then... What's \[\huge \int\limits dt\]
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Very good :) So now, the result of \[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\] is...?
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
1/2 ln y = t + C ?
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
wait question, why is did you put 2y in the denominator in the first place?
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Well, because we want to put everything involving y with the dy and everything involving t with the dt
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
that makes sense... but was my answer correct ?
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Yeah :) But we're not yet done, hang on
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
let's \[\huge \frac{1}{2}\ln(y)=t+C\]playing with it a bit... \[\huge \ln(y) = 2t + 2C\]
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
But 2C is just another constant, so we can disregard the coefficient ln(y) = 2t + C Getting it so far?
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
yup i get it
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
oh then you put e to both sides to get sid of the ln ?
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
That's right :) and e^C is again, just another constant, so just replace it with C And you're done :)
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
how come it's addition and not multiplication ?
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
y=C e ^2t ? that's the answer..
 one year ago

PeterPan Group TitleBest ResponseYou've already chosen the best response.0
Actually, it's \[\huge y = e^{2t + C}\] right? By laws of exponents, this is just \[\huge y = e^Ce^{2t}\] And once again, I said e^C is just another constant anyway, so you can just replace it with C. :)
 one year ago

alyannahere Group TitleBest ResponseYou've already chosen the best response.0
ohhh !!! i see... thank you sooo much !!!!!!!!!
 one year ago
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