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Question
Form the differential equation of all circles which pass through the origin and whose centers lie on X-axis.
Solution
Let C(h, 0) be the center of the circle which passes through the origin. Then the radius of the circle is h.
∴ equation of the circle is (x - h)2 + (y - 0)2 = h2
∴ x2 - 2hx + h + y2 = h2
∴ x2 + h2 = 2hx ....(1)
Differentiating both sides w.r.t. x, we get
`"2x" + "2y" "dy"/"dx" = "2h"`
Substituting the value of 2h in equation (1), we get
`"x"^2 + "y"^2 = ("2x" + "2y" "dy"/"dx")"x"`
∴ `"x"^2 + "y"^2 = 2"x"^2 + 2"xy" "dy"/"dx"`
∴ `"2xy" "dy"/"dx" + "x"^2 - "y"^2 = 0`
This is the required D.E.
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∴ `ye^(x^2) = square`
This is the general solution.
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