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प्रश्न
Find the particular solution of the following differential equation, given that y = 0 when x = `pi/4`.
`(dy)/(dx) + ycotx = 2/(1 + sinx)`
उत्तर
The differential equation is a linear differential equation
IF = `e^(int cotxdx) = e^(logsinx) = sinx`
The general solution is given by
`ysinx = int 2 sinx/(1 + sinx) dx`
⇒ `ysinx = 2 int (sinx + 1 - 1)/(1 + sinx) dx = 2 int [1 - 1/(1 + sinx)] dx`
⇒ `ysinx = 2 int [1 - 1/(1 + cos(pi/2 - x))] dx`
⇒ `ysinx = 2 int [1 - 1/(2cos^2 (pi/4 - x/2))] dx`
⇒ `ysinx = 2 int [1 - 1/2 sec^2 (pi/4 - x/2)] dx`
⇒ `ysinx = 2[x + tan(pi/4 - x/2)] + c`
Given that y = 0, when x = `pi/4`,
Hence, 0 = `2[pi/4 + tan pi/8] + c`
⇒ `c = - pi/2 - 2 tan pi/8`
Hence, the particular solution is `y = "cosec"x [2{x + tan (pi/4 - x/2)} - (pi/2 + 2tan pi/8)]`
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