Nth derivative of the function y=$e^{ax}\sin{bx}$

Question

The function given in the question is $y=e^{ax} \sin(bx)$. The first part of the question asks us to proof the following -

$$ y_n = (a \sec \theta)^n e^{ax} \sin\left(bx + n \tan^{-1}{\frac{b}{a}}\right)$$

I already did this part by differentiating the function w.r.t $x$ once and then twice and then putting $a=r\cos{\phi}$ and $b=r\sin{\phi}$ and then simplifying the equation. After that we can put $r^n=(a\sec{\phi})^n$.

But I am not able to understand how to even start the second part which asks us to proof that $y_{n+1} - 2ay_n + (a^2 + b^2) y_{n-1} = 0$. I think that we can apply the Leibnitz's Theorem after simplifying it to only the derivatives of $y$ and $x$, but I think that's not possible. Even through Leibnitz's Theorem, how can we get the term $y_{n-1}$? Can anyone solve the second part of the problem?

Note - The number in the subscript denotes the number of times $y$ has been differentiated.

https://math.stackexchange.com/questions/1752455/100-th-derivative-of-the-function-fx-ex-cosx/1752462#1752462 — lab bhattacharjee, Sep 19 '19 at 13:45
since you have $$y^{(n)} \text {you have also } y^{(n-1)} and y^{(n+1)} $$ — trula, Sep 19 '19 at 13:48

score 1 · Answer 1 · answered Sep 19 '19 at 13:51

1

A simpler approach is by means of complex numbers.

$$e^{ax}\sin(bx)=\Im(e^{(a+ib)x})$$

so that

$$\left(e^{ax}\sin(bx)\right)^{(n)}=\Im\left((a+ib)^ne^{(a+ib)x}\right).$$

You can obtain the first result by switching to the polar representation of $a+ib$.

The second result is equivalent to

$$(a+ib)^2-2a(a+ib)+(a^2+b^2)=(a+ib-a)^2+b^2=0.$$

answered Sep 19 '19 at 13:51

Can you please explain the second part in detail? Thank you. – Pratham Yadav Sep 19 '19 at 13:54
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@PrathamYadav: it suffices to factor out $(a+ib)^{n-1}e^{(a+ib)x}$ and drop the $\Im$ operator. – Sep 19 '19 at 13:56
1

I think I understood. That's a really great approach. Thanks!!! – Pratham Yadav Sep 19 '19 at 14:05

score 0 · Accepted Answer · answered Sep 19 '19 at 15:18

Set the function $f$:

$$ f(x) = e^{ax}sin(bx)$$

Converting to sum of exponentials:

$$ = e^{ax}\cdot\frac{e^{bix}-e^{-bix}}{2i}$$ $$ = \frac{e^{(a+bi)x}-e^{(a-bi)x}}{2i}$$

The nth derivative is then trivial:

$$ f(x)^{(n} = \frac{(a+bi)^ne^{(a+bi)x}-(a-bi)^ne^{(a-bi)x}}{2i}$$

Converting to trigronometric and exponential functions:

$$ \frac{(re^{\theta i})^n e^{(a+bi)x}-(re^{-\theta i})^ne^{(a-bi)x}}{2i} = $$

$$ \frac{r^ne^{n\theta i} e^{ax}e^{bxi}-r^ne^{-n\theta i}e^{ax}e^{-bxi}}{2i} = $$

$$ r^n \cdot e^{ax}\cdot\frac{e^{i(bx+n\theta)}-e^{-i(bx+n\theta)}}{2i} = $$

$$ r^n \cdot e^{ax}\cdot sin(bx+n\theta) $$

So, the first term of multiplication is wrong, and the other two terms are good.

Thank you very much!!! – Pratham Yadav Sep 19 '19 at 15:28 — Pratham Yadav, Sep 19 '19 at 15:28

Nth derivative of the function y=$e^{ax}\sin{bx}$

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