मराठी

Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0. - Mathematics

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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.

बेरीज
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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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पाठ 9: Differential Equations - Exercise 9.7 [पृष्ठ ४२०]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 9 Differential Equations
Exercise 9.7 | Q 9 | पृष्ठ ४२०

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