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For the differential equation given, find a particular solution satisfying the given condition: (1+x2)dydx+2xy=11+x2;y=0 when x = 1 - Mathematics

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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

योग
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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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अध्याय 9: Differential Equations - Exercise 9.6 [पृष्ठ ४१४]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.6 | Q 14 | पृष्ठ ४१४

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