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

If y=500e7x+600e−7xy = 500{e^{7x}} + 600{e^{ - 7x}} , show that d2ydx2=49y\dfrac{{{d^2}y}}{{d{x^2}}} = 49y .

Explanation

Solution

Hint : Here, we have to prove that the second order differential of the equation y=500e7x+600e−7xy = 500{e^{7x}} + 600{e^{ - 7x}} is equal to 49y. For that we need to differentiate the equation y=500e7x+600e−7xy = 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=500e7x+600e−7xy = 500{e^{7x}} + 600{e^{ - 7x}} - - - - - - - - - - - - - (1)
And we need to prove that d2ydx2=49y\dfrac{{{d^2}y}}{{d{x^2}}} = 49y .
Now, let us differentiate equation (1). Therefore, we get
⇒y=500e7x+600e−7x ⇒dydx=ddx(500e7x)+ddx(600e−7x)   \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 ddx(ax)=addxx\dfrac{d}{{dx}}\left( {ax} \right) = a\dfrac{d}{{dx}}x . Therefore, we get
⇒dydx=500ddx(e7x)+600ddx(e−7x)\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
⇒ddx(eax)=aex\Rightarrow \dfrac{d}{{dx}}\left( {{e^{ax}}} \right) = a{e^x}
Therefore, we get
⇒dydx=7×500e7x−7×600e−7x\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
⇒d2ydx2=7×7×500e7x−(−7)×7×600e−7x ⇒d2ydx2=7×7×500e7x+7×7×600e−7x ⇒d2ydx2=49×500e7x+49×600e−7x   \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
⇒d2ydx2=49(500e7x+600e−7x)\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 49\left( {500{e^{7x}} + 600{e^{ - 7x}}} \right) - - - - - - - - - - - - - - - - - - (3)
Now, we have y=500e7x+600e−7xy = 500{e^{7x}} + 600{e^{ - 7x}}
Therefore, equation (3) becomes
⇒d2ydx2=49y\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 49y
Hence, we have proved that d2ydx2=49y\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. ddxax=axlog⁡a\dfrac{d}{{dx}}{a^x} = {a^x}\log a
II. ddxeax=aex\dfrac{d}{{dx}}{e^{ax}} = a{e^x}
III. ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}