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प्रश्न
Solve yx2dx + e–x dy = 0
उत्तर
yx2dx + e–x dy = 0
e–x dy = – yx2dx
`1/y "d"y = - x^2 "e"^x "d"x`
Integrating on both sides
`int 1/y "d"y = - int x^2 "e"^x "d"x`
log y = `[(x^2) "e"^x - (2x)"e"^x + 2"e"^x] + "c"`
`log y = - "e"^x [x^2 - 2x + 2] + "c"`
`"e"^x (x^2 - 2x + 2) + log y` = c
log y = `- "e"^x [x^2 - 2x + 2] + "c"`
`"e"^x (x^2 - 2x + 2) + log y` = c
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