Find the solution of the differential equation that satisfies the given initial condition. (dP)/(dt)

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]

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