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Question
For the differential equation, find the general solution:
(1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
Solution
The given equation is
(1 + x2) dy + 2xy dx
= cot x dx
⇒ `dy/dx + (2x)/(1 + x^2) y = (cot x)/ (1 + x^2)` ...(1)
Which is a liner equation of the type
Here `P = (2x)/(1 + x^2)`
and `Q = (cot x)/(1 + x^2)`
Now `int P dx = int (2x)/(1 + x^2) dx`
`⇒ log |1 + x^2| = log (1 + x^2)`
[∵ x2 ≥ 0 ⇒ 1 + x2 > 0 ⇒ |1 + x2| = 1 + x2]
∴ `I.F. = e^(int Pdx) = e^(log (1 + x^2)) = 1 + x^2`
∴ The solution is `y.(I.F.) = int Q. (I.F.) dx + C`
⇒ `y (1 + x^2) = int cot x dx + C`
⇒ y (1 + x2) = log |sin x| + C
⇒ y = (1 + x2)-1 log |sin x| + C (1 + x2)-1
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