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SpyIntel [72]
3 years ago
11

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

Mathematics
1 answer:
kati45 [8]3 years ago
6 0

Answer:

P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2

Step-by-step explanation:

Given

\frac{dP}{dt} = \sqrt{Pt

P(1) = 2

Required

The solution

We have:

\frac{dP}{dt} = \sqrt{Pt

\frac{dP}{dt} = (Pt)^\frac{1}{2}

Split

\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}

Divide both sides by P^\frac{1}{2}

\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}

Multiply both sides by dt

\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt

Integrate

\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt

Rewrite as:

\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt

Integrate the left hand side

\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt

\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt

2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt

Integrate the right hand side

2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c

2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c

2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c ---- (1)

To solve for c, we first make c the subject

c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}

P(1) = 2 means

t = 1; P =2

So:

c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}

c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1

c = 2\sqrt 2 - \frac{2}{3}

So, we have:

2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c

2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}

Divide through by 2

P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}

Square both sides

P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2

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