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प्रश्न
Find the particular solution of the differential equation `dy/dx + 2y tan x = sin x` given that y = 0 when x = `pi/3`
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उत्तर
The given differential equation is,
`dy/dx + 2y tanx = sinx` .....(1)
The above is a linear differential equation of the form of `dy/dx + Py = Q`
where P = 2 tan x; Q = sin x
Now, `If = e^(intPdx) = e^(int2tanxdx) = e^(2log sec x) = sec^2 x`
Now, the solution of (1) is given by
`y xx IF = int [Q xx IF]dx + C`
`=> ysec^2 x = int[sin x xx sec^2x] dx + C`
`=> y sec^2 x = intsecx.tan x dx + C`
when `x = pi/3 , y = 3`
0 = 2 + C
C = -2
Particular solution
`ysec^2x = sec x - 2`
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