The family of curves passing through `(0,0)` and satisfying the differential equation `y_2/y_1=1` `("where", y_n=(d^ny)/dx^n)` is (A) `y=k` (B) `y=kx` (C) `y= k(e^x +1)` (C) `y= k(e^x -1)`

Options

A. y = k
B. y = kx
C. `y = k(e^(x) + 1)`
D. `y = k(e^(x)-1)`

Explanation

To solve the problem, we need to find the family of curves that pass through the point (0,0) and satisfy the differential equation given by:

\[ \frac{y_2}{y_1} = 1 \]

where \(y_n = \frac{d^n y}{dx^n}\). This means we can rewrite the equation as:

\[ \frac{d^2y}{dx^2} = \frac{dy}{dx} \]

### Step 1: Rewrite the Differential Equation
From the equation \(\frac{d^2y}{dx^2} = \frac{dy}{dx}\), we can rearrange it to:

\[ \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0 \]

### Step 2: Integrate the Equation
Now we will integrate both sides. Let \(p = \frac{dy}{dx}\). Then, we have:

\[ \frac{dp}{dx} - p = 0 \]

Integrating this, we get:

\[ \int \frac{dp}{p} = \int dx \]

This leads to:

\[ \ln |p| = x + C_1 \]

where \(C_1\) is a constant of integration. Exponentiating both sides gives:

\[ p = e^{x + C_1} = e^{C_1} e^x \]

Let \(k = e^{C_1}\), then:

\[ \frac{dy}{dx} = k e^x \]

### Step 3: Integrate Again
Next, we integrate \(\frac{dy}{dx} = k e^x\):

\[ y = \int k e^x \, dx = k e^x + C_2 \]

where \(C_2\) is another constant of integration.

### Step 4: Apply the Condition (0,0)
Since the curve passes through the point (0,0), we can substitute \(x = 0\) and \(y = 0\):

\[ 0 = k e^0 + C_2 \implies 0 = k + C_2 \implies C_2 = -k \]

### Step 5: Final Equation
Substituting \(C_2\) back into the equation for \(y\):

\[ y = k e^x - k \]

This can be factored as:

\[ y = k(e^x - 1) \]

### Conclusion
Thus, the family of curves that satisfy the given differential equation and pass through the origin (0,0) is:

\[ y = k(e^x - 1) \]

### Answer
The correct option is (D) \(y = k(e^x - 1)\).