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प्रश्न
In the following example verify that the given function is a solution of the differential equation.
`"x"^2 + "y"^2 = "r"^2; "x" "dy"/"dx" + "r" sqrt(1 + ("dy"/"dx")^2) = "y"`
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उत्तर
`"x"^2 + "y"^2 = "r"^2` .....(1)
Differentiating both sides w.r.t. x, we get
`2"x" + 2"y" "dy"/"dx" = 0`
∴ `2"y" "dy"/"dx" = - 2"x"`
∴ `"dy"/"dx" = - "x"/"y"`
∴ `"x" "dy"/"dx" + "r" sqrt(1 + ("dy"/"dx")^2)`
`= "x" (- "x"/"y") + "r" sqrt(1 + (- "x"/"y")^2)`
`= - "x"^2/"y" + "r" sqrt(1 + "x"^2/"y"^2)`
`= - "x"^2/"y" + "r" sqrt(("y"^2 + "x"^2)/"y"^2)`
`= - "x"^2/"y" + "r" sqrt("r"^2/"y"^2)` ....[By (1)]
`= - "x"^2/"y" + "r"^2/"y" = ("r"^2 - "x"^2)/"y"`
∴ `"y"^2/"y" = "y"`
Hence, x2 + y2 = r2 is a solution of the D.E.
`"x" "dy"/"dx" + "r" sqrt(1 + ("dy"/"dx")^2) = "y"`
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