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Solution of the differential equation `cosx dy= y(sinx -y)dx , 0 lt x lt pi/2` (A) `secx=(tanx+c)y` (B) `ysecx=tanx+c` (C) `ytanx=secx+c` (D) `tanx=(secx+c)y` |
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Answer» `dy/dx = (y(sinx - y))/cosx = y tan x - y^2 secx` `1/y^2 dy/dx = 1/y tan x - secx` let `1/y = t` `-1/y^2 dy/dt = dt/dx` now, `-dt/dx = t tan x - secx` `- dt/dx - t tan x = -sec x` `dt/dx + t tan x = sec x ` `IF = e^(int tan x dx)= e^(ln |secx|)= sec x` `t(IF) = int (IF) sec x ` `t * sec x = int(sec^2 x)` `t sec x = int sec^2 x = tan x+ c` `1/y sec x = tan x + c` `secx = y tanx + cy` option 1 is correct |
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