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Solve the following differential equation: (x2 + y2)dx - 2xy dy = 0

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Question

Solve the following differential equation:

(x2 + y2)dx - 2xy dy = 0

Sum
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Solution

(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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Chapter 6: Differential Equations - Exercise 6.4 [Page 203]

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