The solution of the differential equation `(3sin^(2)xcosx) y^(2)dx+2ysin^(3)xdy=sinx dx` (where, C is an arbitrary constant)

Options

A. `2y^(2)sinx=cosx+C`
B. `y^(2)sin^(3)x+cosx=C`
C. `y^(3)sin^(2)x+sinx=C`
D. `y sin x= cos^(2)x+C`

Answer

To solve the differential equation \((3\sin^{2}x\cos x) y^{2}dx + 2ysin^{3}x dy = \sin x dx\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given differential equation:
\[ (3\sin^{2}x\cos x) y^{2}dx + 2ysin^{3}x dy - \sin x dx = 0 \]
This can be rewritten as:
\[ (3\sin^{2}x\cos x) y^{2}dx + (2ysin^{3}x - \sin x) dy = 0 \]

### Step 2: Identifying the Form
We can identify the left-hand side of the equation as a total differential. We want to express it in the form:
\[ dF = 0 \]
where \(F\) is a function of \(x\) and \(y\).

### Step 3: Finding the Function \(F\)
We need to find a function \(F(x, y)\) such that:
\[ \frac{\partial F}{\partial x} = 3\sin^{2}x\cos x \cdot y^{2} \]
and
\[ \frac{\partial F}{\partial y} = 2\sin^{3}x \cdot y - \sin x \]

### Step 4: Integrating with Respect to \(x\)
Integrate \(\frac{\partial F}{\partial x}\) with respect to \(x\):
\[ F(x, y) = \int 3\sin^{2}x\cos x \cdot y^{2} dx \]
Using the substitution \(u = \sin x\), \(du = \cos x \, dx\):
\[ F(x, y) = 3y^{2} \int u^{2} du = 3y^{2} \cdot \frac{u^{3}}{3} + g(y) = y^{2} \sin^{3}x + g(y) \]
where \(g(y)\) is an arbitrary function of \(y\).

### Step 5: Finding \(g(y)\)
Now, differentiate \(F(x, y)\) with respect to \(y\):
\[ \frac{\partial F}{\partial y} = 2y \sin^{3}x + g'(y) \]
Set this equal to the expression we found earlier:
\[ 2y \sin^{3}x + g'(y) = 2y \sin^{3}x - \sin x \]
From this, we find:
\[ g'(y) = -\sin x \]
Integrating with respect to \(y\) gives:
\[ g(y) = -y \sin x + C \]

### Step 6: Final Function
Thus, we have:
\[ F(x, y) = y^{2} \sin^{3}x - y \sin x + C = 0 \]
This can be rearranged to:
\[ y^{2} \sin^{3}x + \cos x = C \]

### Conclusion
The solution of the differential equation is:
\[ y^{2} \sin^{3}x + \cos x = C \]