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Question
Find a particular solution of the differential equation`(x + 1) dy/dx = 2e^(-y) - 1`, given that y = 0 when x = 0.
Solution
The given equation is
`(x + 1) dy/dx = 2e^-y - 1`
⇒ `dy/(2e^-y - 1) = dx/(x + 1)` ....(1)
Integrating, we get `int dy/(2e^-y - 1) = int dx/(x + 1) + C`
⇒ `int dy/(2e^-y - 1) =log |x + 1| + C`
Now, `I = int dy/ (2e^-y - 1) = int e^y/(2 - e^y) dy`
Put ey = t so that ey dy = dt
∴ `I = int dt/(2-t) = - log |2 - t| = - log |2 - e^y|`
From (1), - log |2 - ey|
= log |x + 1| + C ....(2)
When x = 0. y = 0
∴ - log |2 - 1| = log |0 + 1| + C
⇒ - log |1| = log |1| + C
⇒ 0 = 0 + C
⇒ C = 0
Putting in (2), - log |2 ey| = log |x + 1|
⇒ `log |2 - e^y| = log |1/ (x + 1)|`
⇒ `2 e^y = 1/(x + 1)`
⇒ `e^y = 2 - 1/ (x + 1) = (2x + 1)/(x + 1)`
⇒ `y = log |(2x + 1)/ (x + 1)|, x ne -1`
Which is the required solution.
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