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 one year ago
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
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.

This Question is Closed

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

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

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

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

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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

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

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

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

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

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

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

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

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

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

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

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

alyannahere
 one year ago
Best ResponseYou've already chosen the best response.01/2 ln y = t + C ?

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

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

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

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

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

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

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

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

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

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

PeterPan
 one year ago
Best ResponseYou've already chosen the best response.0Actually, 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. :)

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