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Question
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
Solution
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 + 2c1
∴ 2x log y = (log y)2 + c, where c = 2c1
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