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प्रश्न
Solve the following differential equation:
`"y"^3 - "dy"/"dx" = "x"^2 "dy"/"dx"`
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उत्तर
`"y"^3 - "dy"/"dx" = "x"^2 "dy"/"dx"`
∴ `"y"^3 = "dy"/"dx" + "x"^2 "dy"/"dx"`
∴ `"y"^3 = (1 + "x"^2)"dy"/"dx"`
∴ `1/(1 + "x"^2) "dx" = 1/"y"^3`dy
Integrating both sides, we get
`int 1/(1 + "x"^2) "dx" = int "y"^-3`dy
∴ `tan^-1 "x" = "y"^(-2)/-2 + "c"_1`
∴ `tan^-1 "x" = - 1/"2y"^2 + "c"_1`
∴ 2y2 tan-1 x = - 1 + 2c1y2
∴ 2y2 tan-1 x + 1 = cy2, where c = 2c1
This is the general solution.
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