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# Solve the following differential equation: (x2 - y2)dx - 2xy dy = 0 - Mathematics and Statistics

Sum

Solve the following differential equation:

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

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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 + x" "dv"/"dx"

∴ (1) becomes, v + x "dv"/"dx" = ("x"^2 - "v"^2"x"^2)/(- 2"x" ("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")

∴ (- 2"v")/(1 + "v"^2) "dv" = 1/"x" "dx"

Integrating both sides, we get

∴int (- 2"v")/(1 + "v"^2) "dv" = int 1/"x" "dx"

∴ log |1 + "v"^2| = log "x" + log "c"_1

....[because "d"/"dx" (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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