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प्रश्न
In the following example verify that the given function is a solution of the differential equation.
`"y" = "e"^"ax" sin "bx"; ("d"^2"y")/"dx"^2 - 2"a" "dy"/"dx" + ("a"^2 + "b"^2)"y" = 0`
उत्तर
`"y" = "e"^"ax" sin "bx"`
∴ `"dy"/"dx" = "d"/"dx"("e"^"ax" sin "bx")`
`= "e"^"ax" * "d"/"dx" (sin "bx") + sin "bx" * "d"/"dx" ("e"^"ax")`
`= "e"^"ax" xx cos "bx" * "d"/"dx" ("bx") + sin "bx" * "e"^"ax" * "d"/"dx" ("ax")`
`= "e"^"ax" cos "bx" xx "b" + "e"^"ax" sin "bx" xx "a"`
`= "e"^"ax" ("b" cos "bx" + "a" sin "bx")`
and `("d"^2"y")/"dx"^2 = "d"/"dx"["e"^"ax" ("b" cos "bx" + "a" sin "bx")]`
`= "e"^"ax" * "d"/"dx" ("b" cos "bx" + "a" sin "bx") + ("b" cos "bx" + "a" sin "bx")*"d"/"dx"("e"^"ax")`
`= "e"^"ax" ["b" (- sin "bx") * "d"/"dx" ("bx") + "a" cos "bx" * "d"/"dx"("bx")] + ("b" cos "bx" + "a" sin "bx") * "e"^"ax" * "d"/"dx" ("ax")`
`= "e"^"ax" [- "b" sin "bx" xx "b" + "a" cos "bx" xx "b"] + ("b" cos "bx" + "a" sin bx) * "e"^"ax" xx "a"`
`= "e"^"ax" (- "b"^2 sin "bx" + "ab" cos "bx" +"ab" cos "bx" + "a"^2 sin "bx")`
`= "e"^"ax" [("a"^2 - "b"^2) sin "bx" + 2"ab" cos "bx"]`
∴ `("d"^2"y")/"dx"^2 - 2"a" "dy"/"dx" + ("a"^2 + "b"^2)"y"`
`= "e"^"ax" [("a"^2 - "b"^2)] sin "bx" + 2"ab" cos "bx" - 2"a" * "e"^"ax" ("b" cos "bx" + "a" sin "bx") + ("a"^2 + "b"^2) * "e"^"ax" sin "bx"`
`= "e"^"ax" [("a"^2 - "b"^2) sin "bx" + 2"ab" cos "bx" - 2"ab" cos "bx" - 2"a"^2 sin "bx" + ("a"^2 + "b"^2) sin "bx"]`
`= "e"^"ax" [("a"^2 - "b"^2) sin "bx" - ("a"^2 - "b"^2) sin "bx"]`
`= "e"^"ax" xx 0 = 0`
Hence. y = `"e"^"ax"` sin bx is a solution of the D.E.
`("d"^2"y")/"dx"^2 - 2"a" "dy"/"dx" + ("a"^2 + "b"^2)"y" = 0`
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