The equation of the curve satisfying the differential equation `y_2(x^2+1)=2x y_1` passing through the point (0,1) and having slope of tangent at `x=0` as 3 (where `y_2` and `y_1` represent 2nd and 1st order derivative), then (a) `( b ) (c) y=f(( d ) x (e))( f )` (g) is a strictly increasing function (h) `( i ) (j) y=f(( k ) x (l))( m )` (n) is a non-monotonic function (o) `( p ) (q) y=f(( r ) x (s))( t )` (u) has a three distinct real roots (v) `( w ) (x) y=f(( y ) x (z))( a a )` (bb) has only one negative root.

Options

A. `y=f(x)` is a strictly increasing function
B. `y=f(x)` is a non-monoatomic function
C. `y=f(x)` has three distinct real root
D. `y=f(x)` has only one negative root

Explanation

To solve the given differential equation and find the equation of the curve, we will follow these steps:

### Step 1: Write the given differential equation
The differential equation provided is:
\[ y_2(x^2 + 1) = 2xy_1 \]
where \(y_2\) is the second derivative of \(y\) with respect to \(x\) and \(y_1\) is the first derivative.

### Step 2: Rearrange the equation
We can rearrange the equation to isolate \(y_2\):
\[ y_2 = \frac{2xy_1}{x^2 + 1} \]

### Step 3: Divide both sides by \(y_1\)
Dividing both sides by \(y_1\) gives us:
\[ \frac{y_2}{y_1} = \frac{2x}{x^2 + 1} \]

### Step 4: Integrate both sides
Now, we will integrate both sides with respect to \(x\):
\[ \int \frac{y_2}{y_1} \, dx = \int \frac{2x}{x^2 + 1} \, dx \]
The left side integrates to \(\log(y_1)\) and the right side can be solved using substitution:
\[ \int \frac{2x}{x^2 + 1} \, dx = \log(x^2 + 1) + C \]
Thus, we have:
\[ \log(y_1) = \log(x^2 + 1) + C \]

### Step 5: Exponentiate to solve for \(y_1\)
Exponentiating both sides gives:
\[ y_1 = C(x^2 + 1) \]

### Step 6: Use the slope condition
We know that the slope of the tangent at \(x = 0\) is 3, which means:
\[ y_1(0) = 3 \]
Substituting \(x = 0\) into our equation:
\[ y_1(0) = C(0^2 + 1) = C \]
Thus, \(C = 3\), and we have:
\[ y_1 = 3(x^2 + 1) \]

### Step 7: Integrate \(y_1\) to find \(y\)
Now we will integrate \(y_1\) to find \(y\):
\[ y = \int 3(x^2 + 1) \, dx = 3\left(\frac{x^3}{3} + x\right) + C_2 = x^3 + 3x + C_2 \]

### Step 8: Use the point (0, 1)
Since the curve passes through the point (0, 1), we substitute \(x = 0\) and \(y = 1\):
\[ 1 = 0^3 + 3(0) + C_2 \implies C_2 = 1 \]
Thus, the equation of the curve is:
\[ y = x^3 + 3x + 1 \]

### Step 9: Analyze the function
To determine the nature of the function, we can find the derivative \(y'\):
\[ y' = 3x^2 + 3 = 3(x^2 + 1) \]
Since \(3(x^2 + 1) > 0\) for all \(x\), the function \(y\) is strictly increasing.

### Final Result
The equation of the curve is:
\[ y = x^3 + 3x + 1 \]