The equation of the curve satisfying the differential equation ` Y_2 ( x^2 +1) = 4 xy _1`, passing through the point (0, – 4) and having slope of tangent at x = 0 as 4 is:

Options

A. `y=4((x^5)/(5)+x+(2x^3)/(3)-1)`
B. `y=4((x^5)/(5)-x+(3x ^3)/(3)-1)`
C. `y= 4((x^5)/(5)+x+(2x^3)/(3)+1)`
D. none of these

Answer

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

### Step 1: Rewrite the given differential equation
The given differential equation is:
\[ Y_2 (x^2 + 1) = 4xy_1 \]
where \(Y_2\) represents \(\frac{d^2y}{dx^2}\) and \(Y_1\) represents \(\frac{dy}{dx}\). Therefore, we can rewrite it as:
\[ \frac{d^2y}{dx^2} (x^2 + 1) = 4x \frac{dy}{dx} \]

### Step 2: Substitute \(t = \frac{dy}{dx}\)
Let \(t = \frac{dy}{dx}\). Then, we can express \(\frac{d^2y}{dx^2}\) as \(\frac{dt}{dx}\). Substituting this into the equation gives:
\[ \frac{dt}{dx} (x^2 + 1) = 4xt \]

### Step 3: Separate variables
Rearranging the equation, we get:
\[ \frac{dt}{t} = \frac{4x}{x^2 + 1} dx \]

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

### Step 5: Solve for \(t\)
Exponentiating both sides, we find:
\[ t = C(x^2 + 1)^2 \]

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

### Step 7: Integrate to find \(y\)
Now we integrate to find \(y\):
\[ y = \int C(x^2 + 1)^2 dx \]
Expanding \((x^2 + 1)^2\):
\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]
Thus,
\[ y = C \left( \frac{x^5}{5} + \frac{2x^3}{3} + x \right) + K \]

### Step 8: Use initial conditions to find constants
We know that the curve passes through the point \((0, -4)\) and has a slope of 4 at \(x = 0\). Therefore:
1. At \(x = 0\), \(y(0) = -4\):
\[ y(0) = C(0) + K = -4 \implies K = -4 \]
2. The slope at \(x = 0\) is given by:
\[ \frac{dy}{dx}\bigg|_{x=0} = C(0^2 + 1)^2 = C = 4 \]

### Step 9: Substitute back to find the equation of the curve
Substituting \(C = 4\) and \(K = -4\) into the equation for \(y\):
\[ y = 4\left( \frac{x^5}{5} + \frac{2x^3}{3} + x \right) - 4 \]
Simplifying gives:
\[ y = \frac{4x^5}{5} + \frac{8x^3}{3} + 4x - 4 \]

### Final Equation
Thus, the equation of the curve is:
\[ y = \frac{4}{5}x^5 + \frac{8}{3}x^3 + 4x - 4 \]