logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

If \( y = 500e^{\large 7x} + 600e^{\large -7x}\), show that \( \large\frac{d^2y}{dx^2}\normalsize = 49y\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$(e^{\large x})=e^x$
Step 1:
We have $y=500e^{\large 7x}+600e^{\large-7x}$
Differentiating with respect to $x$
$\large\frac{dy}{dx}$$=500.7e^{\large 7x}+600.(-7)e^{\large -7x}$
Step 2:
$\large\frac{d^2y}{dx^2}$$=500\times 7\times 7e^{\large 7x}+600(-7)\times (-7)e^{\large-7x}$
$\quad\;=500e^{\large 7x}.49+600e^{\large -7x}.49$
$\quad\;=49[500e^{\large 7x}+600e^{-7x}]$
$\quad\;=49y$
Hence proved.
answered May 13, 2013 by sreemathi.v
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...