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प्रश्न
If y = `e^(m tan^-1x)` then show that `(1 + x^2) (d^2y)/(dx^2) + (2x - m) (dy)/(dx)` = 0
उत्तर
Given, y = `e^(m tan^-1x)`
To prove `(1 + x^2) (d^2y)/(dx^2) + (2x - m) (dy)/(dx)` = 0
Proof:
Given: y = `e^(mtan^-1x)`
∴ `(dy)/(dx) = e^(mtan^-1x) xx m/(1 + x^2)`
∴ `(dy)/(dx) = (me^(mtan^-1))/(1 + x^2)`
∴ `(d^2y)/(dx^2) = m[(me^(tan^-1x) (1/(1 + x^2))(1 + x^2) - e^(mtan^-1x) (2x))/((1 + x^2)^2)]`
= `me^(mtan^-1x) [(m - 2x)/((1 + x^2)^2)]`
Now, `(1 + x^2) (d^2y)/(dx^2) + (2x - m) (dy)/(dx)`
= `(1 + x^2) xx me^(tan^-1x) ((m - 2x))/(1 + x^2)^2 + (2x - m) xx (me^(mtan^-1))/(1 + x^2)`
= `(me^(mtan^-1x))/(1 + x^2) [m - 2x + 2x - m]`
= `(me^(mtan^-1x))/(1 + x^2) xx 0`
= 0
Hence proved.
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