Additional Mathematics 0606 / Calculus — Integration / 3.833.83: Second-order differential equation of a curve0606/21/M/J/21 — Question 9 · 7 marksMark as done · Save for later · Show all solutionsA curve is such that d2ydx2=sin(6x−π2)\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \sin\left(6x - \dfrac{\pi}{2}\right)dx2d2y=sin(6x−2π). Given that dydx=12\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}dxdy=21 at the point (π4,13π12)\left(\dfrac{\pi}{4}, \dfrac{13\pi}{12}\right)(4π,1213π) on the curve, find the equation of the curve.[7]▸ Answer▸ Official mark scheme← 3.15: Equation of a curve from its derivative3.88: Area between a cubic curve and a horizontal line →