Question

The differential equation \( (1+3x^2y^2+\beta x^2y^4)dx + (2x^3y+2x^3y^3)dy = 0 \) will be exact differential equation if (assuming typos in original question are corrected as shown):

• A. \(\beta = -1/2\)
• B. \(\beta = 3\)
• C. \(\beta = 2\)
• D. \(\beta = 3/2\)

Hint:

The condition for an exact differential equation \(Mdx + Ndy = 0\) is \( M_y = N_x \). This is a direct application, so be careful with partial differentiation. The original problem likely had typos, as \(4x^2y=6x^2y\) is impossible.

Answer 1

The Correct Option is D

Step 1: Identify M and N and state the condition for exactness.
A differential equation of the form \(M(x,y)dx + N(x,y)dy = 0\) is exact if and only if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). From the (corrected) equation: \( M = 1+3x^2y^2+\beta x^2y^4 \) \( N = 2x^3y+2x^3y^3 \)

Step 2: Calculate the partial derivatives. \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (1+3x^2y^2+\beta x^2y^4) = 6x^2y + 4\beta x^2y^3 \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (2x^3y+2x^3y^3) = 6x^2y + 6x^2y^3 \]

Step 3: Equate the partial derivatives and solve for \(\beta\). \[ 6x^2y + 4\beta x^2y^3 = 6x^2y + 6x^2y^3 \] For this equality to hold for all x and y, the coefficients of the corresponding terms must be equal. \[ 4\beta x^2y^3 = 6x^2y^3 \] \[ 4\beta = 6 \] \[ \beta = \frac{6}{4} = \frac{3}{2} \]