(2xy^2+y)dx+(2y^3-x)dy=0
Answers
Answer:
Step-by-step explanation:
Writing the DE as M dx + N dy = 0
where M = 2y^3 + 2y^2 and N = - 3xy^2 - 2xy we note that (M_y - N_x) / M = (9y + 6) / (2y^2 + 2y) [= g(y), say].
[Notation: M_y denotes the partial derivative wrt y etc].
Then we know that v(y) = exp[ - Integral g(y) dy] is an integtating factor.
A straightforward calculation yields v(y) = 1 / [y^3.(y+1)^(3/2)].
Multiplying through by the IF yields the exact DE 2/[y (y+1)^(1/2)] dx - [(2x + 3xy)/(y^2.(y+1)^(3/2)] dy = 0.
Then the solution is given by F(x,y) = c where
F_x = M = 2/[y(y+1)^(1/2)] and
F_y = N = - [(2x + 3xy )/ (y^2.(y+1)^(3/2))].
Then F = Integral [F_x dx]
= 2x / [y (y+1)^(1/2)] + h(y).
Differentiating this partially wrt y we conclude that h'(y) = 0 so that h(y) = c.
General solution 2x /[y.sqrt (y +1) ] + c = 0.
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