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Question
Find the general solution of (1 + tany)(dx – dy) + 2xdy = 0.
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Solution
Given that: (1 + tan y)(dx – dy) + 2xdy = 0
⇒ (1 + tan y)dx – (1 + tan y)dy + 2xdy = 0
⇒ (1 + tan y)dx – (1 + tan y – 2x)dy = 0
⇒ `(1 + tan y) "dx"/"dy" = (1 + tan y - 2x)`
⇒ `"dx"/"dy" = (1 + tan y - 2x)/(1 + tan y)`
⇒ `"dx"/"dy" = 1 - (2x)/(1 + tan y)`
⇒ `"dx"/"dy" + (2x)/(1 + tan y)` = 1
Here, P = `2/(1 + tan y)` and Q = 1
Integrating factor I.F.
= `"e"^(int 2/(1 + tan y) "dy")`
= `"e"^(int (2cosy)/(siny + cosy)"d"y)`
= `"e"^(int (siny + cosy - siny + cosy)/((siny + cosy)) "dy"`
= `"e"^(int(1 + (cosy - siny)/(siny + cosy))"d"y)`
= `"e"^(int 1."d"y) . "e"^(int(cosy - siny)/(siny + cosy)"d"y)`
= `"e"^y . "e"^(log(siny + cosy)`
= `"e"^y . (siny + cos y)`
So, the solution is `x xx "I"."F". = int "Q" xx "I"."F". "d"y + "c"`
⇒ `x . "e"^y (siny + cosy) = int 1 . "e"^y (siny + cosy)"d"y + "c"`
⇒ `x . "e"^y )siny + cosy) = "e"^y . sin y + "c"` .....`[because int x^x "f"(x) + "f'"(x)]"d"x = "e"^x "f"(x) + "c"]`
⇒ `x(siny + cos y) = sin y + "c" . "e"^-y`
Hence, the required solution is `x(siny + cos y) = sin y + "c" . "e"^-y`.
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