Let the solution curve x=x(y), 0<y<2π, of (loge(cosy))2cosydx−(1+3xloge(cosy))sinydy=0 satisfy x(3π)=2loge21. If x(6π)=logem−logen1, m,n coprime, then mn is \_\_\_\_\_\_.
Answer: 12
Solutions · 1
Official solution
With t=lncosy: dydx−lncosy3tanyx=(lncosy)2tany; I.F.=(lncosy)−3, giving xln3(cosy)=−2ln2(cosy)+C.
Using x(π/3)=2ln21: C=0. At y=π/6: x=ln4−ln31, so m=4,n=3, mn=12.