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The general solution of the differential equation `(dy)/(dx) = y tan x - y^(2) sec x` isA. `tan x = (c + sec x)y`B. `sec y = (c + tan y)x`C. `sec x = (c + tan x) y`D. None of these |
Answer» Correct Answer - C We have `(dy)/(dx) = y tan x - y^(2) sec x` `rArr" "(1)/(y^(2))(dy)/(dx)-(1)/(y) tan x = -sec x` Putting `(1)/(y) = v rArr (-1)/(y^(2))(dy)/(dx)=(dv)/(dx)`, we get `(dv)/(dx) + tan x * v = sec x` which is linear diff. equation `I.F. = e^(int tan xdx) = e^(log sec x) = sec x` `therefore" "`The solution is `v sec x = int sec^(2) xdx + c` `rArr" "(1)/(y) sec x = tan x + c` `rArr" "sec x = y (c + tan x)` |
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