Question

Find the solution to the following initial value problem;
dy/dx= -(2xy-3x^2/x^2 + 6y) with y(0)=1

Accepted Answer

You need to multiply by x^2 + 6y  both sides such that:
-3x^2+ 2xy = (x^2 + 6y)(dy)/(dx)
You should use the following notations F(x,y) = 3x^2 - 2xy  and G(x,y) = x^2 + 6y  and you need to check if the equation is exact such that:
(delF(x,y))/(del y) = (delG(x,y))/(del x) 
2x = 2x
Since (delF(x,y))/(del y) = (delG(x,y))/(del x) = 2x , then the equation is exact and you may consider as solution the constant function f(x,y) = c .
f(x,y) = int (2xy - 3x^2)dx
f(x,y) = g(y) + 2y*x^2/2 - 3x^3/3
f(x,y) = g(y) + x^2*y - x^3
You need to differentiate f(x,y) with respect to y to evaluate g(y) such that:
(del f(x,y))/(del y) = (del(g(y) + x^2*y - x^3))/(del y)
(del f(x,y))/(del y) = g'(y) + x^2
You should solve the following equation such that:
g'(y) + x^2 = G(x,y) = x^2 + 6y => g'(y) = 6y => int g'(y) = int 6y dy
g(y) = 6y^2/2 + c => g(y) = 3y^2
Hence, evaluating the solution f(x,y) yields:
f(x,y) =3y^2 + x^2*y - x^3 => 3y^2 + x^2*y - x^3 - c = 0
You need to solve for y the equation above such that:
y_(1,2) = (-x^2+-sqrt(x^4+ 12x^3 + 12c))/6
Hence, evaluating the solutions to the given first order nonlinear differential equation yields y_(1,2) = (-x^2+-sqrt(x^4 + 12x^3 + 12c))/6.

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Find the solution to the following initial value problem; dy/dx= -(2xy-3x^2/x^2 + 6y) with y(0)=1

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