Differential Equation Root of Pt (aka xy)

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Homework Help Overview

The discussion revolves around solving a differential equation of the form dP/dt = √(Pt), with an initial condition P(1) = 2. Participants are exploring methods to approach this problem, particularly focusing on integration techniques and the separability of the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the possibility of using an integrating factor and express confusion regarding the integration of two variables. Some suggest rewriting the equation to facilitate separation of variables, while others question the necessity of substitution.

Discussion Status

The discussion is active, with participants providing insights into the separability of the equation and clarifying misconceptions about integration techniques. There is a mix of interpretations regarding the best approach to take, but no explicit consensus has been reached.

Contextual Notes

Some participants mention constraints related to their familiarity with integration methods, particularly substitution, and express uncertainty about handling integrals involving multiple variables.

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!
 

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