## alyannahere 2 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.

1. PeterPan

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

2. PeterPan

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

3. PeterPan

@alyannahere ?

4. alyannahere

solving it... :)

5. alyannahere

not really sure how.. ?

6. alyannahere

7. 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

8. alyannahere

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

9. PeterPan

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

10. alyannahere

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

11. alyannahere

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

12. alyannahere

i don't know how he got it

13. PeterPan

Hang on, just integrate for now, and tell me what you got :)$\huge \frac{1}{2}\int\limits\frac{dy}{y}=\int\limits dt$

14. alyannahere

1/2 ln y = C ?

15. PeterPan

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

16. alyannahere

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

17. PeterPan

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

18. alyannahere

then ?

19. PeterPan

Then... What's $\huge \int\limits dt$

20. alyannahere

t+C ?

21. PeterPan

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

22. alyannahere

1/2 ln y = t + C ?

23. alyannahere

:o

24. alyannahere

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

25. PeterPan

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

26. alyannahere

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

27. PeterPan

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

28. PeterPan

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

29. PeterPan

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

30. alyannahere

yup i get it

31. alyannahere

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

32. PeterPan

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

33. alyannahere

how come it's addition and not multiplication ?

34. alyannahere

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

35. PeterPan

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. :)

36. alyannahere

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