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