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प्रश्न
Solve the following differential equation:
(x2 + y2)dx - 2xy dy = 0
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उत्तर
(x2 + y2)dx - 2xy dy = 0
∴ 2xy dy = (x2 + y2)dx
∴ `"dy"/"dx" = ("x"^2 + "y"^2)/"2xy"` ....(1)
Put y = vx
∴ `"dy"/"dx" = "v"+ ("xdv")/"dx"`
∴ (1) becomes, v + x`"dv"/"dx" = ("x"^2 + "v"^2"x"^2)/("2x"("vx"))`
∴ `"v + x""dv"/"dx" = (1 + "v"^2)/"2v"`
∴ `"x""dv"/"dx" = (1 + "v"^2)/"2v" - "v" = (1 + "v"^2 - 2"v"^2)/"2v"`
∴ `"x""dv"/"dx" = (1 - "v"^2)/"2v"`
∴ `"2v"/(1 - "v"^2)"dv" = 1/"x" "dx"`
Integrating both sides, we get
`int"2v"/(1 - "v"^2)"dv" = int 1/"x" "dx"`
`- int"- 2v"/(1 - "v"^2)"dv" = int 1/"x" "dx"`
∴ - log |1 - v2| = log x + log c1 ....`[because "d"/"dv" (1 - "v"^2) = - 2"v" and int("f"'("x"))/("f"("x")) "dx" = log |"f"("x")| + "c"]`
∴ `log |1/(1 - "v"^2)| = log "c"_1 "x"`
∴ `log |1/(1 - ("y"^2/"x"^2))| = log "c"_1 "x"`
∴ `log |"x"^2/("x"^2 - "y"^2)| = log "c"_1 "x"`
∴ `"x"^2/("x"^2 - "y"^2) = "c"_1"x"`
∴ `"x"^2 - "y"^2 = 1/"c"_1 "x"`
∴ `"x"^2 - "y"^2 = "cx"`, where c = `1/"c"_1`
This is the general solution.
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