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प्रश्न
Solve the following differential equation:
`"x" "dy"/"dx" + "2y" = "x"^2 * log "x"`
उत्तर
`"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.
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