Step 1: Define M and N and state the condition for exactness.
A differential equation in the form \(M(x,y)dx + N(x,y)dy = 0\) is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).Given the equation:\( M = 1+3x^2y^2+\beta x^2y^4 \)\( N = 2x^3y+2x^3y^3 \)
Step 2: Compute 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: Set the partial derivatives equal and solve for \(\beta\).\[ 6x^2y + 4\beta x^2y^3 = 6x^2y + 6x^2y^3 \]For this to be true for all x and y, the coefficients of like terms must be equal.\[ 4\beta x^2y^3 = 6x^2y^3 \]\[ 4\beta = 6 \]\[ \beta = \frac{6}{4} = \frac{3}{2} \]
Let \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to ______.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: