2020 THREE(e)

Excellence
Question
A curve has the equation y=(3x+2)e−2xy = (3x + 2)e^{-2x}.

Prove that d2ydx2+4dydx+4y=0\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} + 4y = 0.

*You must use calculus and show any derivatives that you need to find when solving this
problem.*
Official Answer
dydx=(3x+2)e−2x⋅(−2)+3e−2x\dfrac{dy}{dx}=(3x+2)e^{-2x}\cdot(-2)+3e^{-2x}
=e−2x[−2(3x+2)+3]=e^{-2x}\left[-2(3x+2)+3\right]
=e−2x(−6x−1)=e^{-2x}(-6x-1)

d2ydx2=−6e−2x−2e−2x(−6x−1)\dfrac{d^2y}{dx^2}=-6e^{-2x}-2e^{-2x}(-6x-1)
=e−2x[−6−2(−6x−1)]=e^{-2x}\left[-6-2(-6x-1)\right]
=e−2x(−6+12x+2)=e^{-2x}(-6+12x+2)
=e−2x(12x−4)=e^{-2x}(12x-4)
=4e−2x(3x−1)=4e^{-2x}(3x-1)

EITHER
d2ydx2+4dydx+4y=0\dfrac{d^2y}{dx^2}+4\dfrac{dy}{dx}+4y=0
LHS=4e−2x(3x−1)+4e−2x(−6x−1)+4e−2x(3x+2)=4e^{-2x}(3x-1)+4e^{-2x}(-6x-1)+4e^{-2x}(3x+2)
=4e−2x[3x−1−6x−1+3x+2]=4e^{-2x}\left[3x-1-6x-1+3x+2\right]
=0=0
== RHS as required

OR
LHS=e−2x(12x−4)+4e−2x(−6x−1)+4e−2x(3x+2)=e^{-2x}(12x-4)+4e^{-2x}(-6x-1)+4e^{-2x}(3x+2)
=e−2x[12x−4+4(−6x−1)+4(3x+2)]=e^{-2x}\left[12x-4+4(-6x-1)+4(3x+2)\right]
=e−2x[12x−4+4(−6x−1)+4(3x+2)]=e^{-2x}\left[12x-4+4(-6x-1)+4(3x+2)\right]
=e−2x[12x−4−24x−4+12x+8]=e^{-2x}\left[12x-4-24x-4+12x+8\right]
=0=0
== RHS as required
Grading Criteria

Achievement (u)

  • Correct expression for dydx\dfrac{dy}{dx}.

Merit (r)

  • Correct expression for d2ydx2\dfrac{d^2y}{dx^2}.

Excellence T1

  • Correct solution with correct derivatives.

Excellence T2

-
Video Explanation
NCEA Level 3 Calculus Differentiation 2020 NZQA Exam - Worked Answers by infinityplusone(starts at 52:29)
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