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प्रश्न
Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.
उत्तर
The given equation is
(1 + e2x)dy + (1 + y2) ex dx = 0
⇒ `dy/(1 + y^2) + e^x/(1 + e^(2x)) dx = 0`
Integrating, `int dy/(1 + y^2) + int e^x/(1 + e^(2x)) dx = C_1`
⇒ `tan^-1 y + I = C_1` ....(1)
Where `I = int e^x/(1 + e^(2x)) dx`
Put ex = t
⇒ ex dx = dt
∴ `I = int dt/(1 + t^2) = tan^-1 t + C_2`
= tan-1 ex + C2
From (1), tan-1 y + tan-1 ex + C2 = C1
or tan-1 y + tan-1 ex = C .....(2)
When x = 0, y = 1,
∴ tan-1 (1) + tan-1 (e0) = C
⇒ tan-1 (1) + tan-1 (1) = C
⇒ `pi/4 + pi/4 = C`
⇒ `C = pi/2`
Putting in (2) `tan^-1 y + tan^-1 e^x = pi/2,`
Which is the required solution.
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