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Find a particular solution of the differential equation(x+1)dydx=2e-y-1, given that y = 0 when x = 0. - Mathematics

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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.

Sum
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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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Chapter 9: Differential Equations - Exercise 9.7 [Page 421]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 9 Differential Equations
Exercise 9.7 | Q 14 | Page 421

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