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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.

Differential Equations
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First one :) \[\huge \frac{dy}{dt}=2y\] Let's multiply both sides by dt, and divide both sides by 2y
\[\huge \frac{dy}{2y}=dt\] Now integrate both sides :)

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Other answers:

solving it... :)
not really sure how.. ?
i'm really sorry.. i don't know how to go about this problem @PeterPan
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
so that is 1/2 1/y = C ?
No... remember \[\huge \int\limits \frac{dx}{x}=\ln(x) + C\]
so it's 1/2 ln y = C ?
my friend said the answer is y=C e ^2t
i don't know how he got it
Hang on, just integrate for now, and tell me what you got :)\[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\]
1/2 ln y = C ?
Well, isn't it though \[\huge \int\limits dx = x + C\]
how did it become dx ? i'm so sorry i just really don't understand this lesson
This is just a reminder :) The integral of 1 is just x (plus a constant)
then ?
Then... What's \[\huge \int\limits dt\]
t+C ?
Very good :) So now, the result of \[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\] is...?
1/2 ln y = t + C ?
:o
wait question, why is did you put 2y in the denominator in the first place?
Well, because we want to put everything involving y with the dy and everything involving t with the dt
that makes sense... but was my answer correct ?
Yeah :) But we're not yet done, hang on
let's \[\huge \frac{1}{2}\ln(y)=t+C\]playing with it a bit... \[\huge \ln(y) = 2t + 2C\]
But 2C is just another constant, so we can disregard the coefficient ln(y) = 2t + C Getting it so far?
yup i get it
oh then you put e to both sides to get sid of the ln ?
That's right :) and e^C is again, just another constant, so just replace it with C And you're done :)
how come it's addition and not multiplication ?
y=C e ^2t ? that's the answer..
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. :)
ohhh !!! i see... thank you sooo much !!!!!!!!!

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