A curve passes through `(1,0)` and satisfies the differential equation `(2x cos y + 3x^2y) dx +(x^3 - x^2 siny - y)dy = 0`

Options

A. The equation of curve is `x^(2) cos y + x^(3)y - y^(2) = 1`
B. The equation of curve is `x^(2) cos y + x^(3)y - (y^(2))/(2) = 1`
C. The equation of normal at (1, 0) is y = 0
D. The equation of tangent at (1, 0) is x = 1

Answer

To solve the given problem step by step, we will start with the differential equation and then integrate it to find the curve that passes through the point (1, 0).

### Step 1: Write down the given differential equation
The differential equation given is:
\[ (2x \cos y + 3x^2 y) \, dx + (x^3 - x^2 \sin y - y) \, dy = 0 \]

### Step 2: Rearrange the equation
We can rearrange the equation to isolate \(dy\):
\[ dy = -\frac{(2x \cos y + 3x^2 y)}{(x^3 - x^2 \sin y - y)} \, dx \]

### Step 3: Check if the equation is exact
To check if the equation is exact, we need to find \(M(x, y)\) and \(N(x, y)\):
- \(M(x, y) = 2x \cos y + 3x^2 y\)
- \(N(x, y) = x^3 - x^2 \sin y - y\)

Now, we compute the partial derivatives:
- \(\frac{\partial M}{\partial y} = -2x \sin y + 3x^2\)
- \(\frac{\partial N}{\partial x} = 3x^2 - 2x \sin y\)

Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact.

### Step 4: Find an integrating factor
Since the equation is not exact, we will look for an integrating factor. In this case, we can try to find an integrating factor that depends only on \(x\) or \(y\).

After testing various forms, we find that multiplying the entire equation by \(1/x^2\) makes it exact:
\[ \frac{2 \cos y}{x} + 3y \, dx + \left(1 - \frac{\sin y}{x} - \frac{y}{x^2}\right) dy = 0 \]

### Step 5: Solve the exact equation
Now we can integrate \(M\) and \(N\) to find a potential function \(F(x, y)\):
1. Integrate \(M\) with respect to \(x\):
\[ F(x, y) = \int (2 \cos y + 3y) \, dx = 2x \cos y + x^3 y + g(y) \]
where \(g(y)\) is an arbitrary function of \(y\).

2. Differentiate \(F\) with respect to \(y\) and set it equal to \(N\):
\[ \frac{\partial F}{\partial y} = -2x \sin y + x^3 - g'(y) \]
Setting this equal to \(N\):
\[ -2x \sin y + x^3 - g'(y) = 1 - \frac{\sin y}{x} - \frac{y}{x^2} \]

3. Solve for \(g'(y)\) and integrate to find \(g(y)\).

### Step 6: Find the constant of integration
Now we have the implicit solution:
\[ F(x, y) = C \]
To find \(C\), we substitute the point (1, 0):
\[ F(1, 0) = 2(1) \cos(0) + (1)^3(0) - \frac{0^2}{2} = 2 \]
Thus, \(C = 2\).

### Step 7: Write the final equation
The final equation of the curve is:
\[ 2x \cos y + x^3 y - \frac{y^2}{2} = 2 \]