Which one of the following curves represents the solution of the initial value problem `dy/dx= 100-y`, where `y (0) = 50`

Options

A.
B.
C.
D.

Answer

To solve the initial value problem given by the differential equation \( \frac{dy}{dx} = 100 - y \) with the initial condition \( y(0) = 50 \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the equation:
\[ \frac{dy}{dx} = 100 - y \]
We can rearrange this to isolate \( dy \) on one side:
\[ dy = (100 - y)dx \]

### Step 2: Separate Variables
Next, we separate the variables by moving all terms involving \( y \) to one side and \( x \) to the other:
\[ \frac{dy}{100 - y} = dx \]

### Step 3: Integrate Both Sides
Now we integrate both sides. The left side requires a logarithmic integration:
\[ \int \frac{dy}{100 - y} = \int dx \]
The left side integrates to:
\[ -\ln |100 - y| = x + C \]
where \( C \) is the constant of integration.

### Step 4: Solve for \( y \)
To solve for \( y \), we first exponentiate both sides to eliminate the logarithm:
\[ |100 - y| = e^{-x - C} \]
This can be rewritten as:
\[ 100 - y = \pm e^{-x - C} \]
For simplicity, we can let \( k = e^{-C} \) (a positive constant), giving us:
\[ 100 - y = ke^{-x} \]
Thus, we can express \( y \) as:
\[ y = 100 - ke^{-x} \]

### Step 5: Apply Initial Condition
Now, we apply the initial condition \( y(0) = 50 \):
\[ 50 = 100 - ke^{0} \]
This simplifies to:
\[ 50 = 100 - k \implies k = 50 \]

### Step 6: Write the Final Solution
Substituting \( k \) back into the equation for \( y \):
\[ y = 100 - 50e^{-x} \]

### Step 7: Conclusion
Thus, the solution to the initial value problem is:
\[ y = 100 - 50e^{-x} \]