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प्रश्न
Solve the following differential equation:
`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`
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उत्तर
`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`
∴ `"dy"/"dx" = ("x"^2 + "2y"^2)/"xy"` ....(1)
Put y = vx. Then `"dy"/"dx" = "v + x" "dv"/"dx"`
∴ (1) becomes, `"v + x" "dv"/"dx" = ("x"^2 + 2"v"^2"x"^2)/("x" * "vx") = (1 + "2v"^2)/"v"`
∴ `"x" "dv"/"dx" = (1 + "2v"^2)/"v" - "v" = (1 + 2"v"^2 - "v"^2)/"v"`
∴ `"x" "dv"/"dx" = (1 + "v"^2)/"v"`
∴ `"v"/(1 + "v"^2) "dv" = 1/"x" "dx"`
Integrating, we get
∴ `int "v"/(1 + "v"^2) "dv" = int 1/"x" "dx"`
∴ `1/2 int "2v"/(1 + "v"^2) "dv" = int1/"x" "dx" + log "c"_1`
∴ `1/2 log |1 + "v"^2| = log |"x"| + log "c"_1`
∴ `log |1 + "v"^2| = 2 log |"x"^2| + 2 log "c"_1^2`
∴ `log |1 + "v"^2| = log |"cx"^2|, "where" "c" = "c"_1^2`
∴ 1 + v2 = cx2
∴ `1 + "y"^2/"x"^2 = "cx"^2`
∴ `("x"^2 + "y"^2)/"x"^2 = "cx"^2`
∴ `"x"^2 + "y"^2 = "cx"^4`
This is the general solution.
Now, y(1) = 0, i.e. when x = 1, y = 0, we get
1 + 0 = c(1)
∴ c = 1
∴ the particular solution is `"x"^2 + "y"^2 = "x"^4`.
Notes
The answer in the textbook is incorrect.
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