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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Solve the following differential equation y log y = (log y-x)dydx

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प्रश्न

Solve the following differential equation y log y = `(log  y - x) ("d"y)/("d"x)`

बेरीज
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उत्तर

y log y = `(log  y - x) ("d"y)/("d"x)`

∴ `("d"x)/("d"y) = (log y - x)/(y log y)`

∴ `("d"x)/("d"y) + x/(y log y) = (logy)/(y log y)`

∴ `("d"x)/("d"y) + (1/(y log y))x = 1/y`

This equation is of the form `("d"x)/("d"y) + "P"x` = Q.

where P = `1/(y log y)` and Q  `1/y`

∴ I.F = `"e"^(int"Pd"y)`

= `"e"^(int 1/(y log y)  "d"y)`

= `"e"^(log(log y))`    ......`[∵ int  ("f'"(x))/("f"(x)) "d"x = log |"f"(x)| + "c"]`

= log y

∴ Solution of the given equation is

x . (I.F.) = `int"Q"("I"."F".)  "d"y + "c"_1`

∴ x . log y = `int 1/y log y  "d"y + "c"_1`

∴ x log y = `int (log y)/y   "dy" + "c"_1`

In R.H.S., put log y = t

∴ `1/y  "d"y` = dt

∴ x log y = `int "t"  "dt" + "c"_1`

∴ x log y =`"t"^2/2 + "c"_1`

∴ 2x log y = t2 + 2c

∴ 2x log y = (log y)2 + c, where c = 2c1 

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पाठ 2.6: Differential Equations - Attempt the following questions III

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