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

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

You will use integration by parts.

- anonymous

could you show me how to do that?

- anonymous

Set u=t and dv=e^tdt

- anonymous

ok

- anonymous

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

- anonymous

alright thanks i'm trying it now

- anonymous

i'm stuck

- anonymous

I'm scanning

- anonymous

alright thanks

- anonymous

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

- anonymous

alright cool

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Just split the factors up using power laws.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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- anonymous

Here's the scan for the first one.

- anonymous

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

- anonymous

i feel like thats not right at all

- anonymous

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

- anonymous

and

- anonymous

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

- anonymous

I added the constant then.

- anonymous

Equate the two.

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- anonymous

haha ya thanks alot

- anonymous

np

- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

refresh the page...sometimes that helps

- anonymous

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- anonymous

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- anonymous

ah

- anonymous

sweet i got it

- anonymous

happy travels )

- anonymous

thanks man i appreciate it

- anonymous

later

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