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प्रश्न
Solve the following differential equation:
`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`
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उत्तर
`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`
∴ `"x"^2 "dy"/"dx" + "xy""dy"/"dx" = "y"^2`
∴ `("x"^2 + "xy")"dy"/"dx" = "y"^2`
∴ `"dy"/"dx" = "y"^2/("x"^2 + "xy")` ......(1)
Put y = vx
∴ `"dy"/"dx" = "v + x" "dv"/"dx"`
∴ (1) becomes, `"v + x""dv"/"dx" = ("v"^2"x"^2)/("x"^2 + "x"*"vx") = "v"^2/(1 + "v")`
∴ `"x""dv"/"dx" = "v"^2/(1 + "v") - "v" = ("v"^2 - "v" - "v"^2)/(1 + "v")`
∴ `"x""dv"/"dx" = (- "v")/(1 + "v")`
∴ `(1 + "v")/"v" "dv" = - 1/"x" "dx"`
Integrating, we get
`int (1 + "v")/"v" "dv" = - int 1/"x" "dx"`
`int (1/"v" + 1)"dv" = - int1/"x" "dx"`
∴ `int 1/"v" "dv" + int 1 "dv" = - int 1/"x" "dx"`
∴ log |v| + v = - log |x| + c
∴ log `|"y"/"x"| + "y"/"x" = - log |"x"| + "c"`
∴ log |y| - log |x| + `"y"/"x"` = - log |x| + c
∴ `"y"/"x" + log |"y"| = "c"`
This is the general solution.
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