Question

Solve

`y log  y dy/dx + x  – log y = 0`

Answer 1

 `y log  y dy/dx + x  – log y = 0`

∴ `dx/dy + 1/(y log y) x= 1/y`

The given equation is of the form `dx/dy + px = Q`

where, `P = 1 /(y logy )and Q = 1/y`

∴ `I.F. = e ^(int^pdy) = e^(int^(1/(ylogy)) dy) = e ^(log | log y|) = log y`

∴ Solution of the given equation is

`x(I.F.) =int Q (I.F.) dy + c_1`

∴ `x.log y =  int 1/y log y  dy + c_1`

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

Differentiating w.r.t. x, we get

`1/y  dy = dt`

∴ `x log y =int t dt + c_1 = t^2/2 + c_1`

∴`x log y =(logy)^2/2 + c_1`

∴ 2x log y = (log y)² + c  …[2c₁ = c]