Question

If ax² + 2hxy + by² = 0, then show that `("d"^2"y")/"dx"^2` = 0

Answer 1

ax² + 2hxy + by² = 0      ....(i)

Differentiating both sides w.r.t. x, we get

`"a"("2x") + "2h" * "d"/"dx" ("xy") + "b"("2y") "dy"/"dx" = 0`

∴ `2"ax" + 2"h" ["x" * "dy"/"dx" + "y"(1)] + 2"by" "dy"/"dx" = 0` 

∴ `2"ax" + 2"hx" "dy"/"dx" + 2"hy" + 2"by" "dy"/"dx" = 0`

∴ `2 "dy"/"dx" ("hx" + "by") = - 2"ax" - 2"hy"`

∴ `2 "dy"/"dx" = (-2("ax" + "hy"))/("hx" + "by")`

∴ `"dy"/"dx" = (- ("ax" + "hy"))/("hx" + "by")`    ....(i)

ax² + 2hxy + by² = 0 

∴ ax² + hxy + hxy + by² = 0

∴ x(ax + hy) + y(hx + by) = 0

∴ y(hx + by) = - x(ax + hy)

∴ `"y"/"x" = (- ("ax" + "hy"))/("hx" + "by")`    ...(ii)

From (i) and (ii), we get

`"dy"/"dx" = "y"/"x"`      ....(iii)

Again, differentiating both sides w.r.t. x, we get

`("d"^2"y")/"dx"^2 = ("x" * "dy"/"dx" - "y" * "d"/"dx" ("x"))/"x"^2`

`= ("x" * ("y"/"x") - "y"(1))/"x"^2`    ....[From (iii)]

`= ("y - y")/"x"^2`

`= 0/"x"^2`

∴ `("d"^2"y")/"dx"^2 = 0`