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Solve the following differential equation: ddydx+(1+x2)tan-1x dy= 0 - Mathematics

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प्रश्न

Solve the following differential equation:

`y"d"x + (1 + x^2)tan^-1x  "d"y`= 0

योग

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(tan1x) = – log y + log C

log (y(tan1x)) + log y = log C

y tan1x = C

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Solution of First Order and First Degree Differential Equations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Ordinary Differential Equations - Exercise 10.5 [पृष्ठ १६१]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 10 Ordinary Differential Equations
Exercise 10.5 | Q 4. (ii) | पृष्ठ १६१

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