Question 237 of 490MediumMCQ
Let y=y(x) solve dxdy−∣A∣=0 for x>0, where A=(ysinx01−1)… (schematic 3×3 determinant from source). If y(π)=π+2, find y(2π). Solutions · 1
Official solution
∣A∣=−xy+2sinx+2, giving dxdy+xy=2sinx+2; I.F.=x.
\[xy = x^2-2x\cos x+2\sin x+C.\]
Using y(π)=π+2: C=0. At x=π/2: y=2π+π4. Answer the question to unlock