Question

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`

Answer 1

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