Match the following
Column I | Column II |
(i) (1 + xy) xdy + (1 – xy) ydx = 0 | [A] – 2x2y – 2xy2 = c |
(ii) (x2 – 4xy– 2y2)dx + (y2 – 4xy – 2x2)dy = 0 | [B] + log = c |
(iii) ey.dx + (xey – 2y) dy = 0 | [C] – x/y = c |
(iv) = | [D] xey – y2 = c |
Text Solution
Verified by Experts(i) [B]; (ii) [A]; (iii) [D]; (iv) [C]
Ans.
(i) [B]
(ii) [A]
(iii) [D]
(iv) [C]
Sol. (i) (1 + xy) xdy + (1 – xy) ydx = 0
(xdy + ydx) + xy (xdy – ydx) = 0
+ 
= 0
dx = 0
+ d log 
= 0
integrating, we get 
+ log 
= c
(ii) x 2 dx + y 2 dy – 4xy dx – 2y 2 dx – 4xydy – 2x 2 dy = 0
x 2 dx + y 2 dy – 2 (2xy dx + x 2 dy)– 2 (2xydy + y 2 dx) = 0
x 2 dx + y 2 dy – 2d (x 2 y) –2d(xy 2 ) = 0
integrating 
– 2x 2 y – 2xy 2 = c
(iii) e y dx + (xe y – 2y) dy = 0
e y dx + xe y /y – 2ydy = 0
d(xe y ) – d(y 2 ) = 0
integrating
xe y – y 2 = c
(iv) 
= d 
= d 
= c
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