Differential Equation Root of Pt (aka xy)

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avant_t7
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Homework Statement


dP/dt=root of (Pt)


Homework Equations


P(1)=2


The Attempt at a Solution


Well, I figure that the integrating factor is 1, since there is no P(x) value, so it's a matter of finding the integral of root(Pt). I can't solve this because the only method I know is substitution, and it doesn't work on this equation.

Thanks guys
 
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avant_t7 said:

Homework Statement


dP/dt=root of (Pt)


Homework Equations


P(1)=2


The Attempt at a Solution


Well, I figure that the integrating factor is 1, since there is no P(x) value, so it's a matter of finding the integral of root(Pt). I can't solve this because the only method I know is substitution, and it doesn't work on this equation.

Thanks guys
this equation is separable. Write sqrt(Pt) as P1/2t1/2.
 
I see that, I guess I'm just confused by finding the integral of two variables. If I substitute u for Pt, and du=(t)dP/dt + P dt...

will that work?
 
You don't need to do a substitution and you aren't going to have any integrals with two variables.

[tex]\frac{dP}{dt} = \sqrt{Pt} = P^{1/2}t^{1/2}[/tex]

Separate to get the following.
[tex]\frac{dP}{P^{1/2}} = t^{1/2} dt[/tex]

Now integrate.
 
shesh you're right, how silly of me.

THanks a lot Mark!