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प्रश्न
Solve the following differential equation:
`y"d"x + (1 + x^2)tan^-1x "d"y`= 0
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उत्तर
`y"d"x + (1 + x^2)tan^-1x "d"y`
Take t = tan–1x
dt `1/(1 + x^2) "d"x`
The equation can be written as
`("d"x)/((1 + x^2)tan^-) = - ("d"y)/y`
`"dt"/"t" = - ("d"y)/y`
Taking Integration on both sides, we get
`int "dt"/"t" = int ("d"y)/y`
log t = – log y + log C
log(tan–1x) = – log y + log C
log (y(tan–1x)) + log y = log C
y tan–1x = C
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