The velocity v of mass m of a rocket at time t is given by the equation :
`m (dv)/(dt)+V (dm)/(dt)=0`,
where 'V' is the constant velocity of emission. If the rocket starts from rest when `t=0` with mass m, prove that :
`v= V log ((m_(0))/(m))`.

Answer

To solve the given differential equation for the velocity \( v \) of a rocket as a function of its mass \( m \), we start with the equation:

\[ m \frac{dv}{dt} + V \frac{dm}{dt} = 0 \]

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate the terms involving \( dv \) and \( dm \):

\[ m \frac{dv}{dt} = -V \frac{dm}{dt} \]

### Step 2: Separating Variables
Next, we separate the variables \( v \) and \( m \):

\[ \frac{dv}{dt} = -\frac{V}{m} \frac{dm}{dt} \]

Now, we can cancel \( dt \) from both sides:

\[ dv = -\frac{V}{m} dm \]

### Step 3: Integrating Both Sides
Now we integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( m \):

\[ \int dv = -V \int \frac{1}{m} dm \]

### Step 4: Performing the Integration
The integral of \( dv \) is simply \( v \), and the integral of \( \frac{1}{m} dm \) is \( \ln |m| \):

\[ v = -V \ln |m| + C \]

### Step 5: Applying Initial Conditions
We know from the problem statement that at \( t = 0 \), the velocity \( v = 0 \) and the mass \( m = m_0 \). We can use these initial conditions to find the constant \( C \):

\[ 0 = -V \ln |m_0| + C \]

Thus,

\[ C = V \ln |m_0| \]

### Step 6: Substituting Back to Find \( v \)
Now we substitute \( C \) back into our equation:

\[ v = -V \ln |m| + V \ln |m_0| \]

This can be simplified using properties of logarithms:

\[ v = V (\ln |m_0| - \ln |m|) = V \ln \left(\frac{m_0}{m}\right) \]

### Final Result
Thus, we have proved that:

\[ v = V \ln \left(\frac{m_0}{m}\right) \]