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प्रश्न
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
उत्तर
`(x + 1) dy/dx -1 = 2e^(-y)`
∴ `(x + 1) dy /dx = 2/e^y + 1`
∴ `(x + 1) dy /dx = ((2+e^y))/e^y `
∴ `e^y /(2+e^y) dy= dx/(1+x)`
Integrating on both sides, we get
`int e^y/(2+e^y) dy = intdx/(1+x)`
∴ log| 2 + ey| = log |1 + x| + log |c|
∴ log |2 + ey| = log |c(1 + x)|
∴ 2 + ey = c (1 + x) ...(i)
When y = 0, x = 1, we have
2 + e0 = c (1 + 1)
∴ 2 + 1 = 2c
∴ c = `3/2`
Substituting c = `3/2` in (i), we get
`2 + e^y = 3/ 2 (1 + x)`
∴ 4 + 2ey = 3 + 3x
∴ 3x - 2ey - 1 = 0, which is the required particular solution.
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