Question 190 of 490MediumMCQ
If dxdy+2ytanx=sinx, 0<x<2π, y(3π)=0, then the maximum value of y(x) is: Solutions · 1
Official solution
I.F. =sec2x. ysec2x=secx+C. Using y(π/3)=0: C=−2.
\[y = \cos x-2\cos^2x = \frac18-2\left(\cos x-\frac14\right)^2.\]
Maximum value =81. Answer the question to unlock