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Question : 80 of 160
Marks:
+1,
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Solution:
(c) Given, differential equation
(1+x2)‌‌‌=em(tan−1x)−y,y(0)=1 ‌‌‌=‌−‌ ⇒‌‌‌+y(‌)=‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) On comparing with Bernoulli's equation
‌+Py=Q ⇒‌‌P=‌,Q=‌∴‌‌IF=e∫Pdx=e∫‌dx=etan−1xSolution
y Eq. (i), we get
y⋅IF=∫Q⋅IFdx+C⇒‌‌y⋅etan−1x=∫‌⋅etan−1xdx+C‌ Put ‌tan−1x=t⇒‌‌‌dx=dt⇒‌‌y⋅etan−1x=∫emt⋅etdt+C ⇒‌‌y⋅etan−1x‌‌=‌+C‌‌=‌+C∵‌‌y‌‌=1,‌ when, ‌x=01⋅e0‌‌=‌+C⇒‌‌‌‌=‌+CC‌‌=1−‌=‌∴‌‌yetan−1x‌‌=‌+‌
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