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Skip in 10
We are asked to solve the differential equation: [ (x^2y - 2xy^2)dx - (x^3 - 3x^2y)dy = 0. ] First, identify ( M(x, y) = x^2y - 2xy^2 ) and ( N(x, y) = -(x^3 - 3x^2y) = -x^3 + 3x^2y ). Check if the equation is exact: [ \frac{\partial M}{\partial y} = x^2 - 4xy, \quad \frac{\partial N}{\partial x} = -3x^2 + 6xy. ] Since (\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}), it is not exact. Look for an integrating factor. Notice that both (M) and (N) are homogeneous of degree 3. Try (y/x = v \Rightarrow y = vx, dy = vdx + xdv). Substitute (y = vx): [ Mdx - Ndy = [(x^2(vx) - 2x(vx)^2)]dx - [(x^3 - 3x^2(vx))(vdx + xdv)]. ] Simplify (Mdx): [ M = x^2(vx) - 2x(vx)^2 = x^3v - 2x^3v^2 = x^3(v - 2v^2), \quad Mdx = x^3(v - 2v^2)dx ] Simplify (Ndy): [ N = -x^3 + 3x^2(vx) = -x^3 + 3x^3v = x^3(-1 + 3v) ] [ dy = vdx + xdv \Rightarrow Ndy = x^3(-1+3v)(vdx + xdv) = x^3(-v + 3v^2)dx + x^4(-1 + 3v) dv ] Form (Mdx - Ndy): [ Mdx - Ndy = x^3(v - 2v^2)dx - [x^3(-v + 3v^2)dx + x^4(-1
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