Find the value of b of the given equation which is exact and then solve it using the value b.
(ye^{2xy}+x) + bxe^{2xy}y^{'} = 0
M = ye^{2x} + x
N = bxe^{2xy}
M_{y} = e^{2xy} + 2xye^{2xy}
N_{x} = be^{2xy} + 2bxye^{2xy}
Since M_{y} = N_{x}, b will equal to 1.
M = ye^{2xy} + x, N = xe^{2xy}
φ_{x} = M, φ_{y} = N
φ_{x} = ye^{2xy} + x
φ = ∫(ye^{2xy}+x)dx = ye^{2xy}/2y + x^{2}/2 + h(y)
φ_{y} = xe^{2xy} + h^{'}(y)
φ_{y} = xe^{2xy}
h'(y) = 0, then h'(y) = c
φ = e^{2xy}/2 +x^{2}/2 = c