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**Quiz-2 / 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.

(ye

^{2xy}+x) + bxe^{2xy}y^{'}= 0M = ye

^{2x}+ xN = 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}= Nφ

_{x}= ye^{2xy}+ xφ = ∫(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