Question

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

Answer 1

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

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