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प्रश्न
For the following differential equation find the particular solution satisfying the given condition:
`(e^y + 1) cos x + e^y sin x. dy/dx = 0, "when" x = pi/6,` y = 0
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उत्तर
`(e^y + 1) cos x + e^y sin x dy/dx = 0`
`e^y.sinx.dy/dx = - (e^y + 1) cosx`
`inte^y/(e^y + 1).dy = - intcosx/sinx. dx`
`log |e^y + 1| = - log |sinx| + log |c|`
`log |e^y + 1| + log |sinx| = log|c|`
`log|(e^y + 1) . sinx| = log |c|`
`(e^y + 1). sinx = c` ...(i)
when `x = pi/6, y = 0`
`(e^0 + 1). sin(pi/6) = 0`
`(1 + 1) . 1/2 = c`
`2 xx 1/2 = c`
c = 1
From (i)
∴ the particular solution is (ey + 1). sinx = 1
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