A particle falls in a medium whose resistance is propotional to the square of the velocity of the particles. If the differential equation of the free fall is `(dv)/(dt) = g-kv^(2)` (k is constant) then

Options

A. `v=2sqrt(g/k)(e^(2tsqrt(g//t))+1)/(e^(2rsqrt(g//k))-1)`
B. `v=sqrt(g/k)(e^(2tsqrt(gk))-1)/(e^(2tsqrt(gk))+1`
C. `v to 0` as `t to infty`
D. `v to sqrt(g/k)` as `t to infty`

Explanation

To solve the differential equation given by \(\frac{dv}{dt} = g - kv^2\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to separate the variables \(v\) and \(t\):
\[ \frac{dv}{g - kv^2} = dt \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side requires a specific integral formula. The integral of \(\frac{1}{a^2 - x^2}\) is given by:
\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C \]
In our case, we set \(a^2 = \frac{g}{k}\) (thus \(a = \sqrt{\frac{g}{k}}\)), and we rewrite the integral:
\[ \int \frac{dv}{g - kv^2} = \int \frac{dv}{\frac{g}{k} - v^2} \]
This leads to:
\[ \frac{1}{\sqrt{gk}} \ln \left| \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v} \right| = t + C \]

### Step 3: Solving for \(v\)
To isolate \(v\), we exponentiate both sides:
\[ \left| \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v} \right| = e^{\sqrt{gk}(t + C)} \]
Let \(C' = e^{\sqrt{gk}C}\), then:
\[ \frac{\sqrt{\frac{g}{k}} + v}{\sqrt{\frac{g}{k}} - v} = C' e^{\sqrt{gk}t} \]
Cross-multiplying gives:
\[ \sqrt{\frac{g}{k}} + v = C' e^{\sqrt{gk}t} \left( \sqrt{\frac{g}{k}} - v \right) \]
Rearranging terms leads to:
\[ v(1 + C' e^{\sqrt{gk}t}) = C' e^{\sqrt{gk}t} \sqrt{\frac{g}{k}} - \sqrt{\frac{g}{k}} \]
Thus:
\[ v = \frac{C' e^{\sqrt{gk}t} - 1}{C' e^{\sqrt{gk}t} + 1} \sqrt{\frac{g}{k}} \]

### Step 4: Finding the Constant of Integration
To find \(C'\), we can use initial conditions. Assuming \(v(0) = 0\):
\[ 0 = \frac{C' - 1}{C' + 1} \sqrt{\frac{g}{k}} \]
This implies \(C' = 1\).

### Final Solution
Substituting \(C' = 1\) back into our equation gives:
\[ v(t) = \frac{e^{\sqrt{gk}t} - 1}{e^{\sqrt{gk}t} + 1} \sqrt{\frac{g}{k}} \]