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प्रश्न
The particular solution of `dy/dx = xe^(y - x)`, when x = y = 0 is ______.
विकल्प
`e^(x - y) = x + 1`
`e^(x + y) = x + 1`
`e^x + e^y = x + 1`
`e^(y - x) = x - 1`
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उत्तर
The particular solution of `dy/dx = xe^(y - x)`, when x = y = 0 is `underlinebb(e^(x - y) = x + 1)`.
Explanation:
`dy/dx = xe^(y - x)`
∴ `int e^-y dy = intxe^-x dx`
∴ `e^-y = x * e^-x/-1 - int 1 * e^-x/-1 dx + c`
∴ `-e^-y = -xe^-x + e^-x/-1 + c`
∴ `e^-y = x/e^x + 1/e^x - c`
∴ `e^(x - y) = x + 1 - ce^x`
When x = y = 0, we get
1 = 1 – c ∴ c = 0
∴ Particular solution is
`e^(x - y) = x + 1`
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