(x^2.y^2 + xy +1)ydx + (x^2.y^2 - xy +1)xdy = 0 Solve the given differential equation with atleast two possible methods.
Question
(x^2.y^2 + xy +1)ydx + (x^2.y^2 - xy +1)xdy = 0
Solve the given differential equation with at least two possible methods.
Solution
The given differential equation is a homogeneous differential equation. We can solve it using two methods:
Method 1: Substitution Method We can make the substitution v = y/x or y = vx. Then, dy = vdx + xdv. Substituting these into the original equation, we get:
(x^2.v^2 + xv +1)vdx + (x^2.v^2 - xv +1)xdv = 0
This simplifies to:
v^3.x^2.dx + (v^2 - v + 1)x.dv = 0
This is a separable equation, which can be solved by separating the variables and integrating.
Method 2: Exact Differential Equation Method We can rewrite the given differential equation as:
Mdx + Ndy = 0, where M = (x^2.y^2 + xy +1)y and N = (x^2.y^2 - xy +1)x
We can check if this equation is exact by checking if ∂M/∂y = ∂N/∂x. If it is, then the solution to the differential equation is given by ∫Mdx + ∫(N - ∂/∂x ∫Mdx)dy = C.
Both methods should give the same solution.
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