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Solve the following differential equation: xydydxx2yyxydydx=x2+2y2,y(1)=0

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Question

Solve the following differential equation:

`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`

Sum
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Solution

`"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`.

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Notes

The answer in the textbook is incorrect.

  Is there an error in this question or solution?
Chapter 6: Differential Equations - Exercise 6.4 [Page 203]

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