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प्रश्न
Find the particular solution of the following differential equation:
y(1 + log x) = (log xx) `"dy"/"dx"`, when y(e) = e2
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उत्तर
`y(1 + log x) = (log x^x) dy/dx`
`y(1 + log x) - (log x^x) dy/dx = 0`
`y(1 + log x)dx/dy - xlogx = 0`
∴ `(1 + log "x")/("x log x")"dx" - "dy"/"y" = 0`
Integrating both sides, we get
∴ `int (1 + log "x")/("x log x")"dx" - "dy"/"y" = "c"_1` .....(1)
Put x log x = t
Then `["x" * "d"/"dx" (log "x") + (log "x") * "d"/"dx" ("x")]"dx" = "dt"`
∴ `["x"/"x" + (log "x")(1)]"dx" = "dt"`
∴ `int (1 + log "x")/("x" log "x")"dx" = int"dt"/"t" = log |"t"| = log |"x" log "x"|`
∴ from (1), the general solution is
log |x log x| - log |y| = log c, where c1 = log c
∴ log `|("x" log "x")/"y"| = log "c"`
∴ `("x" log "x")/"y" = "c"`
∴ x log x = cy
This is the general solution.
Now, y = `"e"^2`, when x = e
∴ e log e = c.e2
∴ 1 = c.e
∴ c = `1/"e"`
∴ the particular solution is x log x = `(1/"e")"y"`
∴ y = ex log x.
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