Question

Find the general solution of the differential equation: `y log y dx/dy+x=2/y`.

Answer 1

Given,

`y log y dx/dy+x=2/y`

`dx/dy+1/(ylogy)x=2/(y^2logy)`

I.F = `e^(intpdy)`

= `e^(int1/(ylogy)dy)`

Let log y = t

`1/ydy=dt`

⇒ dy = ydt

I.F. = `e^(inty/(y.t)dt)`

= `e^(intdt/t)`

= `e^(log|t|)`

= `e^(log|logy|)`

= logy

x × 1·F = ∫(Q × I·F) dy + C

xlogy = `int(2/(y^2logy)*logy)dy+C`

⇒ xlogy = `2inty^-2dy+C`

= `2*(y^-2+1)/(-2+1)+C`

⇒ xlogy = `(-2)/y+C`

⇒ xlogy = `C-2/y`