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

### Step 1: Substitute Variables
Let \( \frac{dy}{dx} = t \). Then, we have:
\[ \frac{d^2y}{dx^2} = \frac{dt}{dx} \]
Substituting these into the original equation gives:
\[ (x^2 + 1) \frac{dt}{dx} = 2x t \]

### Step 2: Rearrange the Equation
Rearranging the equation, we get:
\[ \frac{dt}{dx} = \frac{2x}{x^2 + 1} t \]

### Step 3: Separate Variables
Now, we can separate the variables:
\[ \frac{dt}{t} = \frac{2x}{x^2 + 1} dx \]

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

### Step 5: Exponentiate Both Sides
Exponentiating both sides gives:
\[ t = C (x^2 + 1) \]

### Step 6: Substitute Back for \(t\)
Recall that \(t = \frac{dy}{dx}\), so we have:
\[ \frac{dy}{dx} = C (x^2 + 1) \]

### Step 7: Integrate Again
Integrating both sides with respect to \(x\):
\[ y = \int C (x^2 + 1) dx = C \left( \frac{x^3}{3} + x \right) + D \]
where \(D\) is a constant of integration.

### Step 8: Apply Initial Conditions
Now we apply the initial conditions. First, using \(y'(0) = 3\):
\[ y'(x) = C \left( x^2 + 1 \right) \]
At \(x = 0\):
\[ y'(0) = C(0^2 + 1) = C = 3 \]
So, \(C = 3\).

### Step 9: Substitute \(C\) Back
Substituting \(C\) back into the equation for \(y\):
\[ y = 3 \left( \frac{x^3}{3} + x \right) + D = x^3 + 3x + D \]

### Step 10: Use the Second Initial Condition
Now, using the second condition \(y(0) = 1\):
\[ y(0) = 0^3 + 3(0) + D = D = 1 \]

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