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प्रश्न
For the differential equation given, find a particular solution satisfying the given condition:
`(1 + x^2)dy/dx + 2xy = 1/(1 + x^2); y = 0` when x = 1
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उत्तर
The given equation is
`(1 + x^2) dy/dx + 2xy = 1/(1 + x^2)`
or `dy/dx + (2x)/(1 + x^2) y = 1/(1 + x^2)^2` ....(1)
Which is a linear equation of the type
`dy/dx + Py = Q`
Here `P = (2x)/(1 + x^2)`
and `Q = (1/(1 + x^2))^2`
∴ `int Pdx = int (2x)/(1 + x^2) dx = log |1 + x^2| = log (1 + x^2)`
[∵ x2 ≥ 0 ⇒ 1 + x2 > 0 ⇒ |1 + x2| = 1 + x2]
∴ `I.F. = e^(log (1 + x^2)) = (1 + x^2)`
∴ The solution is `y. (L.F.) = int Q. (I.F.) dx + C`
⇒ `y. (1 + x^2) = int ((1 + x^2))/((1 + x^2)^2) dx + C`
⇒ `y (1 + x^2) = tan^-1 x + C` ....(2)
When x = 1, y = 0,
∴ 0 = tan-1 1 + C
⇒ `C = -pi/4`
Putting in (2), we get `y (1 + x^2) = tan^-1 x - pi/4`
Which is the required solution.
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