Answer: \(y=-\frac54\cos2x+2x+\frac{17\pi}{12}+\frac{5\sqrt3}{8}\).
Given that
\(\frac{d^2y}{dx^2}=5\cos2x,\)
integrate once to find the gradient:
\(\frac{dy}{dx}=\frac52\sin2x+c.\)
The gradient is \(\frac34\) when \(x=-\frac{\pi}{12}\), so
\(\frac34=\frac52\sin\left(-\frac{\pi}{6}\right)+c.\)
Since \(\sin\left(-\frac{\pi}{6}\right)=-\frac12\),
\(\frac34=-\frac54+c.\)
Hence
\(c=2.\)
So
\(\frac{dy}{dx}=\frac52\sin2x+2.\)
Integrate again:
\(y=-\frac54\cos2x+2x+d.\)
The curve passes through
\(\left(-\frac{\pi}{12},\frac{5\pi}{4}\right).\)
Substitute these values:
\(\frac{5\pi}{4} =-\frac54\cos\left(-\frac{\pi}{6}\right)+2\left(-\frac{\pi}{12}\right)+d.\)
Since \(\cos\left(-\frac{\pi}{6}\right)=\frac{\sqrt3}{2}\),
\(\frac{5\pi}{4} =-\frac{5\sqrt3}{8}-\frac{\pi}{6}+d.\)
Therefore
\(d=\frac{5\pi}{4}+\frac{\pi}{6}+\frac{5\sqrt3}{8}.\)
So
\(d=\frac{17\pi}{12}+\frac{5\sqrt3}{8}.\)
The equation of the curve is
\(y=-\frac54\cos2x+2x+\frac{17\pi}{12}+\frac{5\sqrt3}{8}.\)
