Additional Mathematics 0606 / Calculus — Differentiation 2 / 2.542.54: Second derivative identity with a sine combination0606/22/O/N/19 — Question 2 · 5 marksMark as done · Save for later · Show all solutionsGiven that y=2sin3x+cos3xy = 2\sin 3x + \cos 3xy=2sin3x+cos3x, show that d2ydx2+dydx+3y=ksin3x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + \dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y = k\sin 3xdx2d2y+dxdy+3y=ksin3x, where kkk is a constant to be determined.[5]▸ Answer▸ Official mark scheme← 2.27: Exponential product: stationary point, tangent and integral1.37: Cosine curve intersecting a horizontal line →
