MATH 106 Ohio State

Solutions of Selected Problems in HW7 5.2.4 Let y = (y1 , y2 , y3 , y4 )T S . Then < x1 , y >=< x1 , y >= 0 or 1 0 -2 1 0 1 3 -2 y1 y2 y3 y4 = 0 0 . Solving this equation gives the general solution (y1 , y2 , y3 , y4 )T = (2 - , -

Math 150 Final Exam Form B Please be aware that this is a practice exam based on previous department exams. There may be topics from class that are not represented on this practice exam that are important. Please refer to the information given by you

Homework II 2.1.14. y + 2y = te-2t , y(1) = 0 This is a first order linear ODE, (t) = exp 2dt = e2t . Note that the last equality is a slight lie, since it ignores the constant of integration, but any constant will work. e2t (y + 2y) = te-2t e2t = t

Trigonometric Definitions Trigonometric Functions: sin \" ! b r a cos \" ! r b tan \" ! a r b r sec \" ! a a cot \" ! b csc \" ! r b ! a Graphing Trigonometric Functions: y ! A sin #/x % C $ Amplitude ! A y ! A cos( /x % C ) Period ! 2& / P4aseS4if

Topics For Midterm II 3.1 Homogeneous Equations with Constant Coefficients Characteristic Equation to find general solutions, solutions to IVPs. (This works only for linear homogeneous ODEs with constant coefficients, although it can be used as a st