The differential equation:
$$2xydx−(3x^2−y^2)dy=0$$
We are asked to find the value of ( $a$ ) such that ( $y^a$ ) is an integrating factor of the given differential equation.
An integrating factor is a function that, when multiplied by a differential equation, makes it exact. For an equation in the form $( M(x, y)dx + N(x, y)dy = 0 )$, an integrating factor $( \mu(x, y) )$ will satisfy:
$$\frac{\partial}{\partial x}N(x, y) = \frac{\partial}{\partial y}M(x,y)$$
Given $( M(x, y) = 2xy )$ and $( N(x, y) = -(3x^2 - y^2) )$, we need to find $( a )$ such that $( y^a )$ is the integrating factor.
Let’s calculate the partial derivatives:
$$\frac{\partial }{\partial x} -(3x^2-y^2)y^a = -6x^2y^a$$
$$\frac{\partial}{\partial y} 2xy^{a+1} = 2(a+1)xy^a$$
For the equation to be exact, these two partial derivatives should be equal:
$$2(a+1)xy^a=−6xy^a$$
Solving for ( a ):
$$a+1=−3$$
$$a=−4$$
Therefore, the value of ( a ) is -4, corresponding to option (A).
