Tamil Nadu Board of Secondary EducationHSC Commerce Class 11

If y = 500e7x + 600e-7x, then show that y2 – 49y = 0. - Business Mathematics and Statistics

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Sum

If y = 500e7x + 600e-7x, then show that y2 – 49y = 0.

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Solution

y = 500e7x + 600e-7x 

`y_1 = "dy"/"dx" = 500 "d"/"dx" (e^(7x)) + 600 "d"/"dx" (e^(-7x))`

`= 500 (7e^(7x)) + 600(- 7e^(-7x))`

`y_2 = ("d"^2"y")/"dx"^2 = 500xx7 "d"/"dx" (e^(7x)) + 600(-7) "d"/"dx" (e^(-7x))`

`= 500 xx 7(7e^(7x)) + 600 xx (-7)(-7) e^(-7x)`

`= 500 xx 49e^7x + 600 xx 49e^(-7x)`

`y_2 = 49 [500 r^(7x) + 600e^(-7x)]` = 49y

(or) y2 – 49y = 0

Concept: Differentiation Techniques
  Is there an error in this question or solution?
Chapter 5: Differential Calculus - Exercise 5.9 [Page 123]

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