Question

(2xy^2+y)dx+(2y^3-x)dy=0

Answer 1

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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Answer 2

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