Differential Equation Root of Pt (aka xy)

[avant_t7]

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

 

[Mark44]

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 P^1/2t^1/2.

 

[avant_t7]

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?

 

[Mark44]

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

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

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

Now integrate.

 

[avant_t7]

shesh you're right, how silly of me.

THanks a lot Mark!