- 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.

- jamiebookeater

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- anonymous

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

- anonymous

\[\huge \frac{dy}{2y}=dt\]
Now integrate both sides :)

- anonymous

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## More answers

- anonymous

solving it... :)

- anonymous

not really sure how.. ?

- anonymous

i'm really sorry.. i don't know how to go about this problem @PeterPan

- 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

so that is 1/2 1/y = C ?

- anonymous

No... remember
\[\huge \int\limits \frac{dx}{x}=\ln(x) + C\]

- anonymous

so it's 1/2 ln y = C ?

- anonymous

my friend said the answer is y=C e ^2t

- anonymous

i don't know how he got it

- 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

1/2 ln y = C ?

- anonymous

Well, isn't it though
\[\huge \int\limits dx = x + C\]

- anonymous

how did it become dx ?
i'm so sorry i just really don't understand this lesson

- anonymous

This is just a reminder :)
The integral of 1 is just x (plus a constant)

- anonymous

then ?

- anonymous

Then... What's
\[\huge \int\limits dt\]

- anonymous

t+C ?

- anonymous

Very good :)
So now, the result of
\[\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt\]
is...?

- anonymous

1/2 ln y = t + C ?

- anonymous

:o

- anonymous

wait question, why is did you put 2y in the denominator in the first place?

- anonymous

Well, because we want to put everything involving y with the dy
and everything involving t with the dt

- anonymous

that makes sense... but was my answer correct ?

- anonymous

Yeah :)
But we're not yet done, hang on

- anonymous

let's \[\huge \frac{1}{2}\ln(y)=t+C\]playing with it a bit...
\[\huge \ln(y) = 2t + 2C\]

- anonymous

But 2C is just another constant, so we can disregard the coefficient
ln(y) = 2t + C
Getting it so far?

- anonymous

yup i get it

- anonymous

oh then you put e to both sides to get sid of the ln ?

- anonymous

That's right :)
and e^C is again, just another constant, so just replace it with C
And you're done :)

- anonymous

how come it's addition and not multiplication ?

- anonymous

y=C e ^2t ? that's the answer..

- 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

ohhh !!! i see... thank you sooo much !!!!!!!!!

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