The solution of differential equation `(d^2y)/(dx^2)=dy/dx,y(0)=3` and `y'(0)=2` :

Options

A. is a periodic function
B. approaches to zero as `x to -oo`
C. has an asymptote parallel to x-axis
D. has an asymptote parallel to y-axis

Answer

To solve the differential equation \(\frac{d^2y}{dx^2} = \frac{dy}{dx}\) with the initial conditions \(y(0) = 3\) and \(y'(0) = 2\), we will follow these steps:

### Step 1: Rewrite the differential equation
The given equation is:
\[ \frac{d^2y}{dx^2} = \frac{dy}{dx} \]

### Step 2: Integrate the equation
Integrate both sides with respect to \(x\):
\[ \int \frac{d^2y}{dx^2} \, dx = \int \frac{dy}{dx} \, dx \]
This gives:
\[ \frac{dy}{dx} = y + C \]
where \(C\) is a constant of integration.

### Step 3: Apply initial conditions
We know \(y'(0) = 2\) and \(y(0) = 3\). Substitute \(x = 0\):
\[ \frac{dy}{dx}\bigg|_{x=0} = y(0) + C \]
Substituting the known values:
\[ 2 = 3 + C \]
Solving for \(C\):
\[ C = 2 - 3 = -1 \]

### Step 4: Substitute \(C\) back into the equation
Now substitute \(C\) back into the equation:
\[ \frac{dy}{dx} = y - 1 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \frac{dy}{dx} - y = -1 \]
This is a linear first-order differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = -1\) and \(Q(x) = -1\).

### Step 6: Find the integrating factor
The integrating factor \(\mu(x)\) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -1 \, dx} = e^{-x} \]

### Step 7: Multiply through by the integrating factor
Multiply the entire equation by the integrating factor:
\[ e^{-x}\frac{dy}{dx} - e^{-x}y = -e^{-x} \]

### Step 8: Integrate both sides
The left-hand side can be rewritten as:
\[ \frac{d}{dx}(e^{-x}y) = -e^{-x} \]
Integrate both sides:
\[ \int \frac{d}{dx}(e^{-x}y) \, dx = \int -e^{-x} \, dx \]
This results in:
\[ e^{-x}y = e^{-x} + C \]

### Step 9: Solve for \(y\)
Multiply through by \(e^{x}\):
\[ y = 1 + Ce^{x} \]

### Step 10: Apply the initial condition to find \(C\)
Substituting \(x = 0\) and \(y(0) = 3\):
\[ 3 = 1 + Ce^{0} \]
This simplifies to:
\[ 3 = 1 + C \implies C = 2 \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = 1 + 2e^{x} \]