The solution of the differential equation `(x ^(2) +1)(d^(2)y)/(dx ^(2)) =2x (dy)/(dx)` under the conditions y(0)=1 and y '(0) =3,` is

Options

A. `y=x ^(2) +3x +1`
B. ` y= x^(3)+3x +1`
C. `y-x ^(4) +3x +1`
D. ` y=3 tan ^(-1) x+x^(2) +1`

Answer

To solve the differential equation \((x^2 + 1) \frac{d^2y}{dx^2} = 2x \frac{dy}{dx}\) with the initial conditions \(y(0) = 1\) and \(y'(0) = 3\), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ (x^2 + 1) \frac{d^2y}{dx^2} = 2x \frac{dy}{dx} \]
We can rewrite this as:
\[ \frac{d^2y}{dx^2} = \frac{2x}{x^2 + 1} \frac{dy}{dx} \]

### Step 2: Introduce a Substitution
Let \(u = \frac{dy}{dx}\). Then, we have:
\[ \frac{d^2y}{dx^2} = \frac{du}{dx} \]
Substituting this into the equation gives:
\[ \frac{du}{dx} = \frac{2x}{x^2 + 1} u \]

### Step 3: Separate Variables
We can separate the variables:
\[ \frac{du}{u} = \frac{2x}{x^2 + 1} dx \]

### Step 4: Integrate Both Sides
Now, we integrate both sides:
\[ \int \frac{du}{u} = \int \frac{2x}{x^2 + 1} dx \]
The left side integrates to:
\[ \ln |u| = \ln |x^2 + 1| + C_1 \]
For the right side, we can use the substitution \(t = x^2 + 1\), \(dt = 2x dx\):
\[ \int \frac{2x}{x^2 + 1} dx = \ln |x^2 + 1| + C_2 \]
Thus, we have:
\[ \ln |u| = \ln |x^2 + 1| + C \]

### Step 5: Exponentiate to Solve for \(u\)
Exponentiating both sides gives:
\[ u = C(x^2 + 1) \]
Recalling that \(u = \frac{dy}{dx}\), we have:
\[ \frac{dy}{dx} = C(x^2 + 1) \]

### Step 6: Integrate to Find \(y\)
Now we integrate again to find \(y\):
\[ y = \int C(x^2 + 1) dx = C\left(\frac{x^3}{3} + x\right) + C_3 \]

### Step 7: Apply Initial Conditions
We apply the initial conditions \(y(0) = 1\) and \(y'(0) = 3\):
1. From \(y'(0) = 3\):
\[ y'(x) = C(x^2 + 1) \Rightarrow y'(0) = C(0^2 + 1) = C = 3 \]
2. Now substituting \(C = 3\) into the equation for \(y\):
\[ y = 3\left(\frac{x^3}{3} + x\right) + C_3 = x^3 + 3x + C_3 \]
3. From \(y(0) = 1\):
\[ y(0) = 0 + 0 + C_3 = 1 \Rightarrow C_3 = 1 \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = x^3 + 3x + 1 \]