Question

If $ y = 500{e^{7x}} + 600{e^{ - 7x}} $ , show that $ \dfrac{{{d^2}y}}{{d{x^2}}} = 49y $ .

Answer 1

Hint: Here, we have to prove that the second order differential of the equation $ y = 500{e^{7x}} + 600{e^{ - 7x}} $ is equal to 49y. For that we need to differentiate the equation $ y = 500{e^{7x}} + 600{e^{ - 7x}} $ twice and take out the common terms and we will get our answer.

Complete step-by-step answer:
In this question, we are given a function and we need to prove that its second order derivative is equal to 49y.
Given function is: $ y = 500{e^{7x}} + 600{e^{ - 7x}} $ - - - - - - - - - - - - - (1)
And we need to prove that $ \dfrac{{{d^2}y}}{{d{x^2}}} = 49y $ .
Now, let us differentiate equation (1). Therefore, we get
 $
\Rightarrow y = 500{e^{7x}} + 600{e^{ - 7x}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {500{e^{7x}}} \right) + \dfrac{d}{{dx}}\left( {600{e^{ - 7x}}} \right) \;
  $
Now, we know that $ \dfrac{d}{{dx}}\left( {ax} \right) = a\dfrac{d}{{dx}}x $ . Therefore, we get
 $ \Rightarrow \dfrac{{dy}}{{dx}} = 500\dfrac{d}{{dx}}\left( {{e^{7x}}} \right) + 600\dfrac{d}{{dx}}\left( {{e^{ - 7x}}} \right) $
Now, we know that the derivative of
 $ \Rightarrow \dfrac{d}{{dx}}\left( {{e^{ax}}} \right) = a{e^x} $
Therefore, we get
 $ \Rightarrow \dfrac{{dy}}{{dx}} = 7 \times 500{e^{7x}} - 7 \times 600{e^{ - 7x}} $ - - - - - - - - - - - - (2)
This is the first order differential equation.
Now, we need to find the second order differential. So, differentiating equation (2) again w.r.t x, we get
 $
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 7 \times 7 \times 500{e^{7x}} - \left( { - 7} \right) \times 7 \times 600{e^{ - 7x}} \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 7 \times 7 \times 500{e^{7x}} + 7 \times 7 \times 600{e^{ - 7x}} \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 49 \times 500{e^{7x}} + 49 \times 600{e^{ - 7x}} \;
  $
Here, we can take out 49 as common. Therefore, we get
 $ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 49\left( {500{e^{7x}} + 600{e^{ - 7x}}} \right) $ - - - - - - - - - - - - - - - - - - (3)
Now, we have $ y = 500{e^{7x}} + 600{e^{ - 7x}} $
Therefore, equation (3) becomes
 $ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 49y $
Hence, we have proved that $ \dfrac{{{d^2}y}}{{d{x^2}}} = 49y $ .

Note: This was a simple formula based differential question. Below are some important differentials one should keep in mind always.
I. $ \dfrac{d}{{dx}}{a^x} = {a^x}\log a $
II. $ \dfrac{d}{{dx}}{e^{ax}} = a{e^x} $
III. $ \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} $