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# solve differential equation:(2x^(2)y+y^(3))dx+(xy^(2)-3x^(3))dy=0 the solution is: x^2 y^12=c(2 y^2-x^2)^5 but how?

2075 Points
one year ago
hello apurva, we can write given eqn as
dy/dx= – (2x^(2)y+y^(3))/(xy^(2)-3x^(3))
but this is the standard homogeneous form as we can write  – (2x^(2)y+y^(3))/(xy^(2)-3x^(3))=  – (2v+v^(3))/(v^(2)-3) where v= y/x so that y= xv or dy/dx= v + xdv/dx
so, we have v + xdv/dx= – (2v+v^(3))/(v^(2)-3)
or xdv/dx= – (2v+v^(3))/(v^(2)-3) – v= f(v) (say)
so, dv/f(v)= dx/x which is simply the variable separation form, and can be easily solved by simple integration as
∫dv/f(v)= ∫dx/x + C
after you perform the integration, simply substitute v= y/x back into the eqn and u ll get the final ans.
kindly approve :))