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For the differential equation find a particular solution satisfying the given condition: when[xsin2(yx-y)]dx+x dy=0;y=π4when x=1 - Mathematics

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प्रश्न

For the differential equation find a particular solution satisfying the given condition:

`[xsin^2(y/x - y)] dx + x  dy = 0; y = pi/4 "when"  x = 1`

योग
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उत्तर

The given equation is

`[x sin^2 (y/x) - y] dx + x  dy  = 0`

`dy/dx = (x sin^2  y/x - y)/x`

`= (y - x sin^2  y/x)/x`

`= y/x - sin^2  y/x`                  ....(1)

Put y = vx so that `dy/dx = v - sin^2 v`

∴ (1) become: v + x `(dv)/dx = v - sin^2 v`

⇒ `x (dv)/dx = - sin^2 v`

⇒ `cosec^2 v  dv = - dx/x`

Integrating, `int cosec^2 v  dv = - int dx/x`

⇒ - cot v = - log |x| + C

⇒ log |x| - cot v = C

⇒ `log |x| - cot  y/x = C`              ....(2)

When x = 1, `y = pi/4`

∴ `log |1| - cot pi/4 = C`

⇒ 0 - 1 = C

⇒ C = -1

Putting in (2), log |x| - cot `y/x` = -1

⇒ `cot (y/x) = log |x| + log |e|`

Which is the required solution.

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अध्याय 9: Differential Equations - Exercise 9.5 [पृष्ठ ४०६]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.5 | Q 13 | पृष्ठ ४०६

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