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प्रश्न
If y = `log(x + sqrt(x^2 + a^2))^m`, show that `(x^2 + a^2)(d^2y)/(dx^2) + x "d"/"dx"` = 0.
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उत्तर
y = `log(x + sqrt(x^2 + a^2))^m`
= `mlog(x + sqrt(x^2 + a^2))`
∴ `"dy"/"dx" = m"d"/"dx"[log(x + sqrt(x^2 + a^2))]`
= `m xx (1)/(x + sqrt(x^2 + a^2))."d"/"dx"(x + sqrt(x^2 + a^2))`
= `m/(x + sqrt(x^2 + a^2)) xx [1 + (1)/(2sqrt(x^2 + a^2))."d"/"dx"(x^2 + a^2)]`
= `m/(x + sqrt(x^2 + a^2)) xx [1 + (1)/(2sqrt(x^2 + a^2)).(2x + 0)]`
= `m/(x + sqrt(x^2 + a^2)) xx (sqrt(x^2 + a^2) + x)/(sqrt(x^2 + a^2)`
∴ `"dy"/"dx" = m/sqrt(x^2 + a^2)`
∴ `sqrt(x^2 + a^2)"dy"/"dx"` = m
∴ `(x^2 + a^2)(dy/dx)^2` = m2
Differentiating both sides w.r.t. x, we get
`(x^2 + a^2)."d"/"dx"(dy/dx)^2 + (dy/dx)^2."d"/"dx"(x^2 + a^2) = "d"/"dx"(m^2)`
∴ `(x^2 + a^2) xx 2"dy"/"dx"."d"/"dx"(dy/dx) + (dy/dx)^2 xx (2x + 0)` = 0
∴ `(x^2 + a^2) . 2"dy"/"dx"(d^2y)/(dx^2) + 2x (dy/dx)^2` = 0
Cancelling `2"dy"/"dx"` throughtout, we get
`(x^2 + a^2)(d^2y)/(dx^2) + x"dy"/"dx"` = 0.
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