Question

Solve the following differential equation:

`"x" "dy"/"dx" + "2y" = "x"^2 * log "x"`

Answer 1

`"x" "dy"/"dx" + "2y" = "x"^2 * log "x"`

∴ `"dy"/"dx" + (2/"x") * "y" = "x" * log "x"`     ...(1)

This is the linear differential equation of the form

`"dy"/"dx" + "P"*"y" = "Q"`,

where P = `2/"x" and "Q" = "x" * log "x"`

∴ I.F. = `"e"^(int "P dx")`

= `"e"^(int 2/"x" "dx")`

= `"e"^(2 int 1/"x" "dx")`

= `"e"^(2 log "x")`

= `"e"^(log "x"^2)`

= `"x"^2`

∴ The solution of (1) is given by

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

∴ `"y" * "x"^2 = int ("x log x")*"x"^2 "dx" + "c"`

∴ `"x"^2 * "y" = int "x"^3 * log "x  dx" + "c"`

`= (log "x") int "x"^3 "dx" - int ["d"/"dx" (log "x")int "x"^3 "dx"] "dx" + "c"`

`= (log "x") * "x"^4/4 - int 1/"x" * "x"^4/4 "dx" + "c"`

`= 1/4 "x"^4 log "x" - 1/4 int "x"^3 "dx" + "c"`

∴ `"x"^2 * "y" = 1/4 "x"^4 log "x" - 1/4 * "x"^4/4 + "c"`

∴ `"x"^2 * "y" = ("x"^4 log "x")/4 - "x"^4/16 + "c"`

This is the general solution.