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

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

Mathematics
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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\[x=e^t(t-1)+c\]

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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)
1 Attachment
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:\[y^{1/2}=\frac{1}{3}x^{3/2}+c\]
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.
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
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sweet i got it
happy travels )
thanks man i appreciate it
later

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