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प्रश्न
Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2 y dy = 0, give that `y = pi/4` when x = 0
उत्तर
`e^x tan y dx = (e^x - 2)sec^2y dy`
Using variable seprable
`((sec^2y)/(tan y)) dy = (e^x/(e^x - 2))dx`
Integrating both side
`int ((sec^2 y)/(tany)) dy = int (e^x/(e^x - 2)) dx`
Let tan y=p ⇒ sec2y dy = dp and (ex−2) = q ⇒ exdx =dq
`int (dp)/p = int (dq)/q`
In (p) = In (q) + In(c), where c is constant of integration
p = qc
Replacing the values
`tan y = c(x^x - 2)`
When x = 0, `y = pi/4`
1 = c(-1)
c = -1
`tan y = 2 - e^x`
`y = tan^(-1) (2 - e^x)`
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