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Question
If y = `"e"^("m"tan^(-1)x`, show that `(1 + x^2) ("d"^2y)/("d"x^2) + (2x - "m")("d"y)/("d"x)` = 0
Solution
y = `"e"^("m"tan^(-1)x` .......(i)
Differentiating w. r. t. x, we get
`("d"y)/("d"x) = "d"/("d"x)("e"^("m"tan^-1x))`
= `"e"^("m"tan^-1x)*"d"/("d"x)("m"tan^-1x)`
= `"e"^("m"tan^-1x)*"m"/(1 + x^2)`
= `("m"y)/(1 + x^2)` .......[From (i)]
∴ `(1 + x^2)("d"y)/("d"x)` = my
Again, differentiating w. r. t. x, we get
`(1 + x^2)*"d"/("d"x)(("d"y)/("d"x)) + ("d"y)/("d"x)*"d"/("d"x)(1 + x^2) = "m"("d"y)/("d"x)`
∴ `(1 + x^2)*("d"^2y)/("d"x^2) + ("d"y)/("d"x)*2x = "m"("d"y)/("d"x)`
∴ `(1 + x^2) ("d"^2y)/("d"x^2) + (2x - "m")("d"y)/("d"x)` = 0
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