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प्रश्न
Solve the following differential equation:
`"x" sin ("y"/"x") "dy" = ["y" sin ("y"/"x") - "x"] "dx"`
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उत्तर
`"x" sin ("y"/"x") "dy" = ["y" sin ("y"/"x") - "x"] "dx"`
∴ `"dy"/"dx" = ("y" sin ("y"/"x") - "x")/("x" sin ("y"/"x"))` .....(1)
Put y = vx
∴ `"dy"/"dx" = "v + x" "dv"/"dx" and "y"/"x" = "v"`
∴ (1) becomes, v + x`"dv"/"dx" = ("vx sin v - x")/("x sin v")`
∴ `"x" "dv"/"dx" = ("v" sin ("v" - 1))/(sin "v") - "v"`
∴ `"x" "dv"/"dx" = ("v sin v - 1 - v sin v")/(sin "v") = (- 1)/sin "v"`
∴ sin v dv = `- 1/"x"`dx
Integrating both sides, we get
`int sin "v" "dv" = - int 1/"x" "dx"`
∴ - cos v = - log x - c
∴ cos `("y"/"x") = log "x" + "c"`
This is the general solution.
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