Author Topic: TUT0602 Quiz2  (Read 517 times)

Vickyyy

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TUT0602 Quiz2
« on: October 07, 2019, 10:54:47 PM »
Find the value of b of the given equation which is exact and then solve it using the value b.
(ye2xy+x) + bxe2xyy' = 0
M = ye2x + x
N = bxe2xy
My = e2xy + 2xye2xy
Nx = be2xy + 2bxye2xy
Since My = Nx, b will equal to 1.
M = ye2xy + x, N = xe2xy
φx = M, φy = N
φx = ye2xy + x
φ = ∫(ye2xy+x)dx = ye2xy/2y + x2/2 + h(y)
φy = xe2xy + h'(y)
φy = xe2xy
h'(y) = 0, then h'(y) = c
φ = e2xy/2 +x2/2 = c