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प्रश्न
Find the particular solution of the differential equation
`tan x * (dy)/(dx) = 2x tan x + x^2 - y`; `(tan x != 0)` given that y = 0 when `x = pi/2`
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उत्तर
The given differential equation is
`tan x * (dy)/(dx) = 2x tan x + x^2 - y`; `(tan x != 0)`
`=> (dy)/(dx) = 2x + x^2 cotx - y cotx`
`=> dy/dx + (cot x)y = 2x + x^2 cot x`
This is a linear differential equation.
Here, P = cot x, Q = 2x + x2 cot x
:. I.F. = `e^(int P dx) = e^(int cot s dx) = e^(log|sin x|) = sin x`
The general solution of this linear differential equation is given by
y(I.F.) = ∫Q(I.F.)dx + C
`=> y*sinx = int(2x + x^2 cotx) sinx dx + C`
`=>y*sinx = int 2xsin x dx + int x^2 cos x dx + C`
`y*sinx = int2x sin x dx + x^2 sinx - int 2 xsin x + C` (Applying integration by parts in the 2nd integral)
`=>y*sinx = x^2 sinx +C`......1
When y = 0, `x = pi/2` (Given)
`:. 0 xx sin pi/2 = pi^2/4 sin pi/4 + C`
`=> C = - pi^2/4`
Substituting the value of C in (1), we get
`ysinx = x^2 sin x - pi^2/4`
`=> (x^2- y) sin x = pi^2/4`
This is the particular solution of the given differential equation
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