given that y = 1 when x = π, find y in terms of x for the differential equation, dy/dx = xycos(x)

y^-1 dy = xcos(x) dx∫y^-1dy = ∫xcos(x) dx ln(y) = ∫xcos(x) dx [using integration by parts to integrate the right hand side] therefore, ln(y) = xsin(x) - ∫sin(x) dxln(y) = xsin(x) + cos(x) + cat y = 1, x = π, therefore, ln(1) = πsin(π) + cos(π) + c0 = 0 - 1 + c therefore, c = 1hence ln(y) = xsin(x) + cos(x) + 1finally, y = e^{xsin(x) + cos(x)+1}