Find the particular solution of the following differential equation:
when x = 0.
The given differential equation is
… (1)
Separating given differential equation, we get
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On integrating, we get
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Let 2 – ey = t
- ey dy = dt
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⇒ - log (2 – ey) = log (x + 1) + C ... (2)
Putting y = 0, when x = 0 in (2), we get
⇒ - log (2 – 1) = log (0 + 1) + C
⇒ 0 = 0 + C
∴ C = 0
Equation (2) becomes - log (2 – ey) = log (x + 1)
⇒ log (x + 1) + log (2 – ey) = 0
⇒ log (x + 1) (2 – ey) = 0
⇒ (x + 1) (2 – ey) = e0 = 1
∴ The solution is (x + 1) (2 – ey) = 1
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