anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
First one :) \[\huge \frac{dy}{dt}=2y\] Let's multiply both sides by dt, and divide both sides by 2y
anonymous
  • anonymous
\[\huge \frac{dy}{2y}=dt\] Now integrate both sides :)
anonymous
  • anonymous
@alyannahere ?

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

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