The solution of `(d v)/(dt)+k/m v=-g` is (a) `( b ) (c) v=c (d) e^(( e ) (f) (g) k/( h )g( i ) (j) t (k))( l )-( m )(( n ) mg)/( o ) k (p) (q) (r)` (s) (b) `( t ) (u) v=c-( v )(( w ) mg)/( x ) k (y) (z) (aa) e^(( b b ) (cc) (dd) k/( e e ) m (ff) (gg) t (hh))( i i ) (jj)` (kk) (c) `( d ) (e) v (f) e^(( g ) (h) (i) k/( j ) m (k) (l) t (m))( n )=c-( o )(( p ) mg)/( q ) k (r) (s) (t)` (u) (d) `( v ) (w) v (x) e^(( y ) (z) (aa) k/( b b ) m (cc) (dd) t (ee))( f f )=c-( g g )(( h h ) mg)/( i i ) k (jj) (kk) (ll)` (mm)

Options

A. `v=ce^(-k/mt)-(mg)/k`
B. `v=c-(mg)/ke^(-k/mt)`
C. `ve^(-k/mt)= c-(mg)/k`
D. `ve^(k/mt)=c-(mg)/k`

Explanation

To solve the differential equation

\[ \frac{dv}{dt} + \frac{k}{m}v = -g, \]

we will follow these steps:

### Step 1: Rearranging the equation
We can rearrange the equation to isolate the derivative on one side:

\[ \frac{dv}{dt} = -\frac{k}{m}v - g. \]

### Step 2: Identifying the integrating factor
This is a linear first-order differential equation. The integrating factor \( \mu(t) \) is given by:

\[ \mu(t) = e^{\int \frac{k}{m} dt} = e^{\frac{k}{m}t}. \]

### Step 3: Multiplying through by the integrating factor
We multiply the entire differential equation by the integrating factor:

\[ e^{\frac{k}{m}t} \frac{dv}{dt} + e^{\frac{k}{m}t} \frac{k}{m} v = -g e^{\frac{k}{m}t}. \]

### Step 4: Recognizing the left-hand side as a derivative
The left-hand side can be expressed as the derivative of a product:

\[ \frac{d}{dt} \left( e^{\frac{k}{m}t} v \right) = -g e^{\frac{k}{m}t}. \]

### Step 5: Integrating both sides
Now we integrate both sides with respect to \( t \):

\[ \int \frac{d}{dt} \left( e^{\frac{k}{m}t} v \right) dt = \int -g e^{\frac{k}{m}t} dt. \]

The left-hand side simplifies to:

\[ e^{\frac{k}{m}t} v = -g \int e^{\frac{k}{m}t} dt. \]

The integral on the right-hand side is:

\[ -g \cdot \frac{m}{k} e^{\frac{k}{m}t} + C, \]

where \( C \) is the constant of integration.

### Step 6: Solving for \( v \)
Now we can express \( v \):

\[ e^{\frac{k}{m}t} v = -\frac{mg}{k} e^{\frac{k}{m}t} + C. \]

Dividing through by \( e^{\frac{k}{m}t} \):

\[ v = -\frac{mg}{k} + Ce^{-\frac{k}{m}t}. \]

### Step 7: Final expression
Thus, the general solution of the differential equation is:

\[ v(t) = C e^{-\frac{k}{m}t} - \frac{mg}{k}. \]