NCEA Calculus 2017 THREE(e - Solution & Explanation

Question

For the function

y = e^x \cos kx

:

(i) Find

\dfrac{dy}{dx}

and

\dfrac{d^2y}{dx^2}

.

(ii) Find all the value(s) of

k

such that the function

y = e^x \cos kx

satisfies the equation

\dfrac{d^2y}{dx^2} - 2\dfrac{dy}{dx} + 2y = 0

for all values of

x

.

Official Answer

(i)

\dfrac{dy}{dx}=e^x\cdot\cos kx+e^x(-k\sin kx)

=e^x(\cos kx-k\sin kx)

\dfrac{d^2y}{dx^2}=e^x(\cos kx-k\sin kx)

\quad\quad +e^x(-k\sin kx-k^2\cos kx)

=e^x(\cos kx-2k\sin kx-k^2\cos kx)

(ii)

\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0.

\Rightarrow e^x(\cos kx-2k\sin kx-k^2\cos kx)

\quad -2e^x(\cos kx-k\sin kx)+2e^x\cos kx=0

\Rightarrow e^x(\cos kx-k^2\cos kx)=0

e

^x\cos kx(1-k^2)=0

k=\pm1

Grading Criteria

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2017 NZQA Exam - Worked Answers by infinityplusone(starts at 62:43)

2017 THREE(e)