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Answer:
[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]
Step-by-step explanation:
Given
[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]
[tex]P(1) = 2[/tex]
Required
The solution
We have:
[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]
[tex]\frac{dP}{dt} = (Pt)^\frac{1}{2}[/tex]
Split
[tex]\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}[/tex]
Divide both sides by [tex]P^\frac{1}{2}[/tex]
[tex]\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}[/tex]
Multiply both sides by dt
[tex]\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt[/tex]
Integrate
[tex]\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]
Rewrite as:
[tex]\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt[/tex]
Integrate the left hand side
[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt[/tex]
[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]
[tex]2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]
Integrate the right hand side
[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c[/tex]
[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c[/tex]
[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex] ---- (1)
To solve for c, we first make c the subject
[tex]c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}[/tex]
[tex]P(1) = 2[/tex] means
[tex]t = 1; P =2[/tex]
So:
[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}[/tex]
[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1[/tex]
[tex]c = 2\sqrt 2 - \frac{2}{3}[/tex]
So, we have:
[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex]
[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}[/tex]
Divide through by 2
[tex]P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}[/tex]
Square both sides
[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]