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प्रश्न
For the differential equation find a particular solution satisfying the given condition:
(x + y) dy + (x – y) dx = 0; y = 1 when x = 1
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उत्तर
given (x + y) dy + (x – y) dx = 0
`=> dy/dx = (y - x)/(y + x)` ...(i)
∵ The powers of the numerator and denominator are the same so this is a homogeneous differential equation.
∴ Putting y = vx
`dy/dx = v + x (dv)/dx` ...(in equation (i))
`=> v + x (dv)/dx = (vx - x)/(vx + x)`
`=> x (dv)/dx = (v - 1)/(v + 1) - v`
`x (dv)/dx = (v - 1 - v^2 - 1)/(v + 1)`
`= - (v^2 + 1)/(v + 1)`
`(v + 1)/(v^2 + 1)dv = - 1/x dx`
On integrating,
`=> 1/2 int (2v)/(v^2 + 1)dv + int 1/(v^2 + 1) dv = - int 1/x dx`
`1/2 log (v^2 + 1) + tan^-1 (v) = - log x + C`
log (v2 + 1) + 2 tan-1 (v) = - 2 log x + 2C
So on putting `y/x` in place of v,
`log ((y^2 + x^2)/x^2) + 2 tan^-1 (y/x) = - log x^2 + 2C`
`log (x^2 + y^2) - log x^2 + 2 tan^-1 (y/x) = - log x^2 x + 2C`
`log (x^2 + y^2) + 2 tan^-1 (y/x) = 2C` ....(ii)
Given y = 1 and x = 1
log (12 + 12) + 2 tan-1 (1) = 2C
log 2 + 2 tan-1 (1) = 2C
2C = log 2 + 2 `xx pi/4 = log 2 + pi/2`
Putting this value of C in equation (ii),
`log (x^2 + y^2) + 2 tan^-1 (y/x) = pi/2 + log 2`
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