Question

y dx – x dy + log x dx = 0

Answer 1

y dx – x dy + log x dx = 0

y dx – x dy = - log x dx

Dividing throughout by dx, we get

`y-x dy/dx =  – log x `

∴ `-xdy/dx + y = - log x`

∴ `dy/dx - 1/(x y) = logx/x`

The given equation is of the form

`dy/dx + py = Q`

where, `P = -1/x and Q = logx/x`

∴ I.F. = `e ^(int^(pdx) = e^(int^(-1/xdx) e ^-logx`

= `e^(logx ^-1) =  x ^-1 = 1/x`

∴ Solution of the given equation is

`y(I.F.) =int Q (I.F.) dx + c`

∴ `y/x = int logx/x xx1/xdx+c`

In R. H. S., put log x = t …(i)

∴ x = e^t

Differentiating (i) w.r.t. x, we get

`1/xdx = dt`

∴ `y/x = int t/e^t dt +c`

∴ `y/x = int te^t  dt +c`

= `t int e^-t dt - int (d/dt(t)xxint e^-t dt) dt +c `

= `-te^-t - int (-e^-t) dt +c`

= `-te^-t + int e^-t dt +c`

= – te^–t – e ^–t + c

= `(-t-t)/e^t + c`

= `(- logx -1)/x +c`

∴ y = cx – (1 + log x)