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anonymous
 3 years 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.
anonymous
 3 years 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.

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anonymous
 3 years 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not really sure how.. ?

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

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so that is 1/2 1/y = C ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it's 1/2 ln y = C ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i don't know how he got it

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

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

anonymous
 3 years 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then... What's \[\huge \int\limits dt\]

anonymous
 3 years 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...?

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

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that makes sense... but was my answer correct ?

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

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

anonymous
 3 years 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?

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

anonymous
 3 years 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 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how come it's addition and not multiplication ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y=C e ^2t ? that's the answer..

anonymous
 3 years 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. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohhh !!! i see... thank you sooo much !!!!!!!!!
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