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प्रश्न
Solve the following differential equation:
(x2 + 3xy + y2)dx - x2 dy = 0
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उत्तर
(x2 + 3xy + y2)dx - x2 dy = 0
∴ x2 dy = (x2 + 3xy + y2)dx
∴ `"dy"/"dx" = ("x"^2 + 3"xy" + "y"^2)/"x"^2` ....(1)
Put y = vx
∴ `"dy"/"dx" = "v + x" "dv"/"dx"`
∴ (1) becomes, `"v + x" "dv"/"dx" = ("x"^2 + 3"x" * "vx" + "v"^2"x"^2)/"x"^2`
∴ `"v + x" "dv"/"dx" = 1 + 3"v" + "v"^2`
`"x" "dv"/"dx" = "v"^2 + 2"v" + 1 = ("v + 1")^2`
∴ `1/("v + 1")^2 "dv" = 1/"x" "dx"`
Integrating, we get
`int ("v + 1")^-2 "dv" = int 1/"x" "dx"`
∴ `("v + 1")^-1/-1 = log |"x"| + "c"_1`
∴ `- 1/("v + 1") = log |"x"| + "c"_1`
∴ `- 1/("y"/"x" + 1) = log |"x"| + "c"_1`
∴ `- "x"/("y + x") = log |"x"| + "c"_1`
∴ `log |"x"| + "x"/("x + y") = - "c"_1`
∴ `log |"x"| + "x"/("x + y") = "c"`, where c = - c1
This is the general solution.
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