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प्रश्न
Reduce the following differential equation to the variable separable form and hence solve:
`("x - y")^2 "dy"/"dx" = "a"^2`
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उत्तर
`("x - y")^2 "dy"/"dx" = "a"^2` .....(1)
Put x - y = u
∴ x - u = y
∴ 1 - `"du"/"dx" = "dy"/"dx"`
∴ (1) becomes, `"u"^2 (1 - "du"/"dx") = "a"^2`
∴ `"u"^2 - "u"^2 "du"/"dx" = "a"^2`
∴ `"u"^2 - "a"^2 = "u"^2 "du"/"dx"`
∴ dx = `"u"^2/("u"^2 - "a"^2)`du
Integrating both sides, we get
`int "dx" = int (("u"^2 - "a"^2) + "a"^2)/("u"^2 - "a"^2)`du
∴ x = `int 1 "du" + "a"^2 int "du"/("u"^2 - "a"^2) + "c"_1`
`1/"2a" log |("u - a")/("u + a")| + "c"_1`
∴ x = x - y + `"a"/2 log |("x - y - a")/("x - y + a")| + "c"_1`
∴ - c1 + y = `"a"/2 log |("x - y - a")/("x - y + a")|`
∴ - 2c1 + 2y = a log `|("x - y - a")/("x - y + a")|`
∴ c + 2y = a log `|("x - y - a")/("x - y + a")|`, where c = - 2c1
This is the general solution.
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