Differential EquationsEasyThe Wronskian & Variation of Parameters ShareThe Wronskian & Variation of Parameters · Easy problemThe equation y′′−2y′+2y=0y''-2y'+2y=0y′′−2y′+2y=0 has solutions y1=etcosty_1=e^{t}\cos ty1=etcost and y2=etsinty_2=e^{t}\sin ty2=etsint. Evaluate their Wronskian at t=ln3t=\ln 3t=ln3.Check Show answer Walk through it step by step Practice more like this Review this concept