TUT0602 Quiz2

[Vickyyy]

Find the value of b of the given equation which is exact and then solve it using the value b.

(ye^2xy+x) + bxe^2xyy^' = 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 + x²/2 + h(y)

φ_y = xe^2xy + h^'(y)

φ_y = xe^2xy

h'(y) = 0, then h'(y) = c

φ = e^2xy/2 +x²/2 = c