Find the constant of intergration by the general solution of the differential equation `(2x^(2)y-2y^(4))dx+(2x^(3)+3xy^(3))dy=0` if curve passes through (1,1).

Answer

To solve the differential equation \( (2x^{2}y - 2y^{4})dx + (2x^{3} + 3xy^{3})dy = 0 \) and find the constant of integration given that the curve passes through the point (1, 1), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ (2x^{2}y - 2y^{4})dx + (2x^{3} + 3xy^{3})dy = 0 \]

### Step 2: Divide by \( x^{3}y \)
To simplify, we divide the entire equation by \( x^{3}y \):
\[ \frac{2x^{2}y}{x^{3}y}dx - \frac{2y^{4}}{x^{3}y}dx + \frac{2x^{3}}{x^{3}y}dy + \frac{3xy^{3}}{x^{3}y}dy = 0 \]
This simplifies to:
\[ \frac{2}{x}dx - \frac{2y^{3}}{x^{3}}dx + \frac{2}{y}dy + \frac{3y^{2}}{x^{2}}dy = 0 \]

### Step 3: Rearrange the Terms
Rearranging gives us:
\[ \left( \frac{2}{x} - \frac{2y^{3}}{x^{3}} \right)dx + \left( \frac{2}{y} + \frac{3y^{2}}{x^{2}} \right)dy = 0 \]

### Step 4: Identify the Integrating Factors
We can express the equation in terms of differentials:
\[ \frac{2}{x}dx + \frac{2}{y}dy + \left( -\frac{2y^{3}}{x^{3}}dx + \frac{3y^{2}}{x^{2}}dy \right) = 0 \]

### Step 5: Integrate Each Term
Integrating the first two terms:
\[ \int \frac{2}{x}dx + \int \frac{2}{y}dy = 2\ln|x| + 2\ln|y| + C \]
For the third term, we can recognize:
\[ -\frac{2y^{3}}{x^{3}}dx + \frac{3y^{2}}{x^{2}}dy = d\left( \frac{y^{3}}{x^{2}} \right) \]
Thus, we have:
\[ d\left( \frac{y^{3}}{x^{2}} \right) = 0 \]

### Step 6: Combine the Integrals
Combining all parts, we have:
\[ 2\ln|x| + 2\ln|y| + \frac{y^{3}}{x^{2}} = C \]

### Step 7: Solve for the Constant of Integration
Now we need to find the constant \( C \) using the point (1, 1):
\[ 2\ln|1| + 2\ln|1| + \frac{1^{3}}{1^{2}} = C \]
Since \( \ln(1) = 0 \):
\[ 0 + 0 + 1 = C \]
Thus, \( C = 1 \).

### Final Result
The constant of integration is:
\[ \boxed{1} \]