Given: (x + 1)^2 dy/dx = 1 + y + xy. If the DE can be changed to dy/dx + P (x) y

Question

Given: (x + 1)^2 dy/dx = 1 + y + xy. If the DE can be changed to dy/dx + P (x) y = f (x), then P (x) = (A) (x + 1)^2: (B) (x + 1): (C) 1/(x + 1)^2: (D) -1/(x + 1): (E) None of these. The integrating factor for the given DE is (A) (x + 1): (B) e^(x + 1): (C) 1/(x + 1): (D) 1/(x + 1)^2: (E) None of these. (e^integral p (x) dx) (dy/dx) + (e^integral p (x) dx) (p (x) y) = (A) y'(e^integral p (x) dx): (B) p (x) y': (C) [p (x) y]': (D) [y e^integral p (x) dx]: (E) None of these. The general solution of the given DE is y = (B) ln |x + 1| + c: (B) (ln |x + 1|)/(x + 1) + x: (C) (ln|x + 1|)/(x + 1) + c/(x + 1): (D)-1/[2 (x + 1)] + c (x + 1): (E) None of these.

Explanation / Answer

8)

p(x)=-1(x+1)

option d

9) integrating factor is option b

e^(x+y)

10) the final equation will be option d

11) general solution is

option c

lnIX+1I)/(X+1)+C/(X+1)

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