Step 1: Understanding the Problem:
The given equation is a first-order differential equation: \( M(x,y)dx + N(x,y)dy = 0 \). The process begins by checking if it's exact. If it's not, an integrating factor is sought to make it exact.
Step 2: Exactness Check and Integrating Factor:
Define \( M = xy^3 + y \) and \( N = 2x^2y^2 + 2x + 2y^4 \).
Calculate partial derivatives:
\[ \frac{\partial M}{\partial y} = 3xy^2 + 1 \]\
\[ \frac{\partial N}{\partial x} = 4xy^2 + 2 \]\
Since \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \), the equation is not exact. Find an integrating factor. Consider:
\[ \frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = \frac{(4xy^2 + 2) - (3xy^2 + 1)}{xy^3 + y} = \frac{xy^2 + 1}{y(xy^2 + 1)} = \frac{1}{y} \]\
This is a function of \(y\) alone, so the integrating factor \( \mu(y) \) is:
\[ \mu(y) = e^{\int \frac{1}{y} dy} = e^{\ln|y|} = y \]\
Step 3: Solve the Exact Equation:
Multiply the original equation by \( y \):
\[ y(xy^3 + y)dx + y(2x^2y^2 + 2x + 2y^4)dy = 0 \]\
\[ (xy^4 + y^2)dx + (2x^2y^3 + 2xy + 2y^5)dy = 0 \]\
This equation is now exact. Let \( M' = xy^4 + y^2 \) and \( N' = 2x^2y^3 + 2xy + 2y^5 \). Solve \( F(x,y) = C \) by integrating \( M' \) with respect to \( x \):
\[ F(x,y) = \int (xy^4 + y^2) dx = \frac{x^2y^4}{2} + xy^2 + g(y) \]\
Find \( g(y) \) by differentiating \( F \) with respect to \( y \) and setting it equal to \( N' \):
\[ \frac{\partial F}{\partial y} = \frac{x^2}{2}(4y^3) + 2xy + g'(y) = 2x^2y^3 + 2xy + g'(y) \]\
Comparing with \( N' = 2x^2y^3 + 2xy + 2y^5 \), we get:
\[ g'(y) = 2y^5 \]\
\[ g(y) = \int 2y^5 dy = \frac{2y^6}{6} = \frac{y^6}{3} \]\
The general solution is \( \frac{x^2y^4}{2} + xy^2 + \frac{y^6}{3} = C_1 \).
Multiply by 6:
\[ 3x^2y^4 + 6xy^2 + 2y^6 = 6C_1 \]\
Let \( C = 6C_1 \). The solution is \( 3x^2y^4 + 6xy^2 + 2y^6 = C \).
Step 4: Final Answer:
The solution matches option (D), with the constant possibly included in the expression. The solution is \( 3x^2y^4 + 6xy^2 + 2y^6 = C \).