If any differentisl equation in the form
`f(f_(1)(x,y)d(f_(1)(x,y)+phi(f_(2)(x,y)d(f_(2)(x,y))+....=0`
then each term can be intergrated separately.
For example,
`intsinxyd(xy)+int((x)/(y))d((x)/(y))=-cos xy+(1)/(2)((x)/(y))^(2)+C`
The solution of the differential equation
`(xy^(4)+y)dx-xdy=0` is

Options

A. `(x^(3))/(4)+(1)/(2)((x)/(y))^(2)=C`
B. `(x^(4))/(4)+(1)/(3)((x)/(y))^(3)=C`
C. `(x^(4))/(4)-(1)/(2)((x)/(y))^(3)=C`
D. `(x^(4))/(4)-(1)/(2)((x)/(y))^(2)=C`

Answer

To solve the differential equation \( (xy^4 + y)dx - xdy = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (xy^4 + y)dx - xdy = 0 \]
We can rearrange it as:
\[ (xy^4 + y)dx = xdy \]

### Step 2: Dividing by \( y^4 \)
Next, we divide the entire equation by \( y^4 \):
\[ \frac{(xy^4 + y)}{y^4}dx = \frac{x}{y^4}dy \]
This simplifies to:
\[ \left( x + \frac{1}{y^3} \right)dx = \frac{x}{y^4}dy \]

### Step 3: Rearranging Terms
We can rearrange the equation to isolate the differentials:
\[ \left( x + \frac{1}{y^3} \right)dx - \frac{x}{y^4}dy = 0 \]

### Step 4: Expressing in Terms of Total Derivative
We can express the equation in terms of total differentials:
\[ d\left( \frac{x}{y} \right) = 0 \]
This implies that \( \frac{x}{y} \) is a constant.

### Step 5: Integrating
Now, we can integrate each term separately:
\[ \int (xy^4)dx + \int ydx - \int xdy = 0 \]

### Step 6: Performing the Integrations
1. The first integral:
\[ \int xy^4 dx = \frac{x^2y^4}{2} \]
2. The second integral:
\[ \int y dx = yx \]
3. The third integral:
\[ \int x dy = xy \]

Combining these gives:
\[ \frac{x^2y^4}{2} + yx - xy = C \]

### Step 7: Final Form
The final solution can be expressed as:
\[ \frac{x^2y^4}{2} + C = 0 \]

### Conclusion
Thus, the solution of the differential equation \( (xy^4 + y)dx - xdy = 0 \) is:
\[ \frac{x^2y^4}{2} + C = 0 \]