Solve the following differential equation ylogydxdy+x = log y - Mathematics and Statistics

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Sum

Solve the following differential equation

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

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Solution

`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]

  Is there an error in this question or solution?
Chapter 1.8: Differential Equation and Applications - Q.5

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