You learnt Exact Differential Equations before?
They are in the form of: M(x,y) dx + N(x,y) dy = 0.
Let M = 4x + xy^2, N =
y + yx^2
Partial diff M w.r.t. y to get 2xy
Partial diff N w.r.t. x to get 2xy as well.
Therefore, this proves that the equation is exact.
Let a function f(x,y) = k such that:
(del f / dx) = M = 4x + xy^2 -
(1)
(del f / dy) = N = y +
yx^2 - (2)
Integrate (del f / dx) w.r.t. x to get:
f(x,y) = 2x^2 + 0.5x^2(y^2) + g(y) - (3)
Partial differentiate f(x,y) w.r.t. y to get
0 + yx^2 + g'(y)
Compare this with (2), you will see that
g'(y) = y
g(y) = 0.5y^2 + c
Sub g(y) into (3)
f(x,y) = 2x^2 + 0.5x^2(y^2) + 0.5y^2 + c
k - c = 2x^2 + 0.5x^2(y^2) + 0.5y^2
C = 2x^2 + 0.5x^2(y^2) + 0.5y^2 where C = k -
c
When x = 1, y = 2,
C = 6
6 = 2x^2 + 0.5x^2(y^2) + 0.5y^2
12 = 4x^2 + x^2(y^2) + y^2
16 = 4 + 4x^2 + x^2(y^2) + y^2
16 = (4+y^2)(1+x^2)
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And I just realised that my method is too unorthodox. Lol!
Follow it if you want to. Cheers! =D