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For the differential equation find a particular solution satisfying the given condition: dydx- yx+cosec(yx)=0;y=0 when x = 1

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प्रश्न

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

`dy/dx -  y/x + cosec (y/x) = 0; y = 0` when x = 1

योग
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उत्तर

Given differential equation

`dy/dx - y/x + cosec (y/x)` = 0              ....(i)

`dy/dx = y/x - cosec (y/x)`

Clearly, this equation is a differential equation.

∴ Putting y = vx

`v + x (dv)/dx` = v - cosec  v in equation (i)

`=> x  (dv)/dx` = - cosec v

sin v dv = `- 1/x` dx

On integrating,

`- int sin  v  dv = int 1/x dx`

cos v = log x + C

So on putting `(y/x)` in place of v,

`cos (y/x) = log |x| + C`     ....(ii)

Given y = 0 if x = 1              ...[from equation (ii)]

cos 0 = log 1 + C

⇒ C = 1

Putting this value of C in equation (ii),

`cos (y/x) = log |x| + log |e|`

`cos (y/x) = log  |ex|`

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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 14 | पृष्ठ ४०६

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