CBSE Class 12 Math 2018 Solved Paper

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Question : 19 of 29
 
Marks: +1, -0
Find the particular solution of the differential equation ex tan y dx + (2−ex)sec2 y dy = 0 given that y = π4 when x = 0
OR
Find the particular solution of the differential equation dydx + 2y tan x = sin x, given that y = 0 when x = π3
Solution:
ex tan y dx + (2−ex)sec2 y dy
∫ ex(ex−2) dx = ∫ sec2ytany dy
now, ∫ f′(x)f(x) dx = log |f (x)| + c
log |(ex−2)| - log |tan y| + log |c| = 0
log |c(ex−2)tany| = 0
c(ex−2)tany = e0 = 1
tan y = (c (ex - 2))
now, x = 0 , y = π4
1 = (x (1 - 2))
c = - 1
so
tan y = (- 1 (ex - 2))
y = tan−1 (2 - ex)
OR
dydx + 2y tan x = sin x
comparing with the standard form dydx + Py = Q
P = 2 tan x , Q = sin x
Now,
I.F. = e∫Pdx = e∫2tanxdx = e2log|secx| = sec2 x
y sec2 x = sec x + c
0 = 2 + c ⇒ c = - 2 (Since when x = π3 , y = 0)
y sec2 x = sec x - 2
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