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प्रश्न
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
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उत्तर
`y log y ("d"x)/("d"y) + x` = log y
∴ `("d"x)/("d"y) + 1/(ylogy) x = 1/y`
The given equation is of the form `("d"x)/("d"y) + "P"x` = Q
where, P =`1/(ylogy)` and Q = `1/y`
∴ I.F. = `"e"^(int^("Pd"y)`
= `"e"^(int^(1/(ylogy) "d"y)`
= `"e"^(log |log y|)`
= log y
∴ Solution of the given equation is
x(I.F.) = `int "Q"("I.F.") "d"y + "c"_1`
∴ x.logy = `int 1/y log y "d"y + "c"_1`
In R. H. S., put log y = t
Differentiating w.r.t. x, we get
`1/y "d"y` = dt
∴ x log y = `int "t" "dt" + "c"_1`
= `"t"^2/2 + "c"_1`
∴ x log y = `(log y)^2/2 + "c"_1`
∴ 2x log y = (log y)2 + c ......[2c1 + c]
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