## anonymous 5 years ago how do you do this separable equations problem: dx/dt = te^t x(0)=1

1. anonymous

$\frac{dx}{dt}=te^t \rightarrow dx=te^tdt \rightarrow \int\limits_{}{}dx=\int\limits_{}{}te^tdt$

2. anonymous

$\frac{dx}{dt}=te^t \rightarrow dx=te^tdt \rightarrow \int\limits_{}{}dx=\int\limits_{}{}te^tdt$

3. anonymous

This is just$x=e^t(t-1)+c$

4. anonymous

To find c, you have$x(0)=e^0(0-1)+c=1 \rightarrow c=2$

5. anonymous

So your equation is$x(t)=e^t(t-1)+2$

6. anonymous

how did you get e^t(t-1) is it necessary to use the product rule for that integration or what?

7. anonymous

You will use integration by parts.

8. anonymous

could you show me how to do that?

9. anonymous

Set u=t and dv=e^tdt

10. anonymous

ok

11. anonymous

I'll write it and scan - it will be quicker.

12. anonymous

alright thanks i'm trying it now

13. anonymous

i'm stuck

14. anonymous

I'm scanning

15. anonymous

alright thanks

16. anonymous

Just have to wait a min, since the software for my scanner is a pain

17. anonymous

alright cool

18. anonymous

do you think you could help me out with another one after this?

19. anonymous

Maybe - I'm supposed to be getting ready for uni.

20. anonymous

alright well its dy/dx = (xy)^(1/2) and i just wasn't sure how you set that one up

21. anonymous

Just split the factors up using power laws.

22. anonymous

so you would end up with dy/sqrt(y) = sqrt x dx

23. anonymous

$\frac{dy}{dx}=(xy)^{1/2} = x^{1/2}y^{1/2} \rightarrow \frac{dy}{y^{1/2}}=x^{1/2}dx$

24. anonymous

then after you integrate would you have 2 sqrt y = 2/3 x^(3/2)

25. anonymous

26. anonymous

Here's the scan for the first one.

27. anonymous

and when it's all said and done i had $y= (x ^{3/2}/3)^2 + (c/2)^2$

28. anonymous

i feel like thats not right at all

29. anonymous

$\int\limits{}{}\frac{dy}{y^{1/2}}=\frac{y^{1/2}}{1/2}=2y^{1/2}$

30. anonymous

and

31. anonymous

$\int\limits_{}{}x^{1/2}dx = \frac{x^{3/2}}{3/2}=\frac{2}{3}x^{3/2}+c$

32. anonymous

33. anonymous

Equate the two.

34. anonymous

$2y^{1/2}=\frac{2}{3}x^{3/2}+c$

35. anonymous

Divide by 2 to clean up:$y^{1/2}=\frac{1}{3}x^{3/2}+c$

36. anonymous

You don't have to write c/2 since c is just a constant - constants just keep absorbing constants since in the end, their value is determined by initial conditions and you'll end up with the same number.

37. anonymous

ohhhh ok i think i got it since c is a constant i dont need to divide it

38. anonymous

haha ya thanks alot

39. anonymous

np

40. anonymous

i really appreciate it

41. anonymous

become a fan then :)

42. anonymous

how?

43. anonymous

there should be a link next to my name in this area we're typing in. It's a blue link that says, "Become a fan"

44. anonymous

Should have a 'thumbs up' icon...

45. anonymous

ya i dont see it?

46. anonymous

it just has hero and the 110 fans

47. anonymous

refresh the page...sometimes that helps

48. anonymous

it just has hero and the 110 fans

49. anonymous

yeah, the site's awkward

50. anonymous

ah

51. anonymous

sweet i got it

52. anonymous

happy travels )

53. anonymous

thanks man i appreciate it

54. anonymous

later