Question

(x-y)²dx + 2xydy=0
in homogeneous method
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Answer 1

Answer:

x dx + 2 x y dy - y^2 dx = 0

We have the differential d(y^2/x) = (2x y dy - y^2 dx)/x^2

Divide differential equation by x^2

Then dx/x + d(y^2/x) = 0

Integrate ln x + y^2/x = C

x ln x + y^2 = C x

Answer 2

Step-by-step explanation:

x−y2)dx+2xydy=0

This equation is in the form M(x,y)dx+N(x,y)dy=0

Equation is exact if ∂M∂y=∂N∂x

M(x,y)=x−y2N(x,y)=2xy⟶⟶∂M∂y=∂N∂x=−2y2y⎫⎭⎬⎪⎪∂M∂y≠∂N∂x

Find integrating factor:

∂M∂y−∂N∂xN(x,y)=−2y−2y2xy=−2x⟶μ=e∫−2xdx=e−2lnx=1x2

(1x−y2x2)dx+2yxdy=0