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प्रश्न
Solve: `("d"y)/("d"x) + "e"^x + y"e"^x = 0`
उत्तर
`("d"y)/("d"x) + "e"^x + y"e"^x = 0`
`("d"y)/("d"x) = - y"e"^x - "e"^x`
`("d"y)/("d"x) = - "e"^x (y + 1)`
`1/((y + 1)) "d"y = - "e"^x "d"x`
Integrating on both sides
`int 1/((y + 1)) "d"x = - int "e"^x "d"x`
⇒ log (y + 1) = `- "e"^x + "c"`
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