Question

The solution of the differential equation \( (xy^3 + y)dx + (2x^2y^2 + 2x + 2y^4)dy = 0 \) is:

• A. \( 3xy^2 + 6y^4x - 2y^6 + C \), where C is an arbitrary constant
• B. \( 3xy^4 + 3xy^2 + y^6 + C \), where C is an arbitrary constant
• C. \( 6xy^2 - 2y^4x + C \), where C is an arbitrary constant
• D. \( 3x^2y^4 + 6xy^2 + 2y^6 + C \), where C is an arbitrary constant

Hint:

For equations of the form \(Mdx + Ndy = 0\), if they are not exact, always check the two standard tests for integrating factors: \( \frac{1}{N}(M_y - N_x) \) and \( \frac{1}{M}(N_x - M_y) \). If one of these yields a function of a single variable, you can easily find the integrating factor.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given equation is a first-order differential equation of the form \( M(x,y)dx + N(x,y)dy = 0 \). We first check if it is exact. If not, we try to find an integrating factor to make it exact.

Step 2: Check for Exactness and Find Integrating Factor:
Let \( M = xy^3 + y \) and \( N = 2x^2y^2 + 2x + 2y^4 \). Calculate the partial derivatives: \[ \frac{\partial M}{\partial y} = 3xy^2 + 1 \] \[ \frac{\partial N}{\partial x} = 4xy^2 + 2 \] Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact. Let's look for an integrating factor. Consider the expression: \[ \frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = \frac{(4xy^2 + 2) - (3xy^2 + 1)}{xy^3 + y} = \frac{xy^2 + 1}{y(xy^2 + 1)} = \frac{1}{y} \] Since this is a function of \(y\) alone, the integrating factor \( \mu(y) \) is given by: \[ \mu(y) = e^{\int \frac{1}{y} dy} = e^{\ln|y|} = y \]
Step 3: Solve the Exact Equation:
Multiply the original equation by the integrating factor \( y \): \[ y(xy^3 + y)dx + y(2x^2y^2 + 2x + 2y^4)dy = 0 \] \[ (xy^4 + y^2)dx + (2x^2y^3 + 2xy + 2y^5)dy = 0 \] This equation is exact. Let the new \( M' = xy^4 + y^2 \) and \( N' = 2x^2y^3 + 2xy + 2y^5 \). The solution \( F(x,y) = C \) is found by integrating \( M' \) with respect to \( x \): \[ F(x,y) = \int (xy^4 + y^2) dx = \frac{x^2y^4}{2} + xy^2 + g(y) \] To find \( g(y) \), differentiate \( F \) with respect to \( y \) and set 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 \( \frac{x^2y^4}{2} + xy^2 + \frac{y^6}{3} = C_1 \). To match the options, multiply the entire equation 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), likely with the constant on one side being represented as part of the expression in the option. A more standard form is \( 3x^2y^4 + 6xy^2 + 2y^6 = C \).