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

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

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

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

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

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

You will use integration by parts.

could you show me how to do that?

Set u=t and dv=e^tdt

ok

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

alright thanks i'm trying it now

i'm stuck

I'm scanning

alright thanks

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

alright cool

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

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

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

Just split the factors up using power laws.

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

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

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

Here's the scan for the first one.

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

i feel like thats not right at all

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

and

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

I added the constant then.

Equate the two.

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

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

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

haha ya thanks alot

np

i really appreciate it

become a fan then :)

how?

Should have a 'thumbs up' icon...

ya i dont see it?

it just has hero and the 110 fans

refresh the page...sometimes that helps

it just has hero and the 110 fans

yeah, the site's awkward

ah

sweet i got it

happy travels )

thanks man i appreciate it

later