Question

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

Answer 1

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

Answer 2

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

Answer 3

This is just\[x=e^t(t-1)+c\]

Answer 4

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

Answer 5

So your equation is\[x(t)=e^t(t-1)+2\]

Answer 6

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

Answer 7

You will use integration by parts.

Answer 8

could you show me how to do that?

Answer 9

Set u=t and dv=e^tdt

Answer 10

ok

Answer 11

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

Answer 12

alright thanks i'm trying it now

Answer 13

i'm stuck

Answer 14

I'm scanning

Answer 15

alright thanks

Answer 16

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

Answer 17

alright cool

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Just split the factors up using power laws.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

Here's the scan for the first one.

Answer 26

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

Answer 27

i feel like thats not right at all

Answer 28

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

Answer 29

and

Answer 30

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

Answer 31

I added the constant then.

Answer 32

Equate the two.

Answer 33

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

Answer 34

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

Answer 35

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.

Answer 36

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

Answer 37

haha ya thanks alot

Answer 38

np

Answer 39

i really appreciate it

Answer 40

become a fan then :)

Answer 41

how?

Answer 42

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"

Answer 43

Should have a 'thumbs up' icon...

Answer 44

ya i dont see it?

Answer 45

it just has hero and the 110 fans

Answer 46

refresh the page...sometimes that helps

Answer 47

it just has hero and the 110 fans

Answer 48

yeah, the site's awkward

Answer 49

ah

Answer 50

sweet i got it

Answer 51

happy travels )

Answer 52

thanks man i appreciate it

Answer 53

later