1 )y h(t) = e t(C 1 C 2t) For the particular solution, (Undet Coefs), y p= Acos(3t) Bsin(3t) Substitute to get (A 6B 9A)sin(3t) (B 6A 9B)cos(3t #color(blue)("Some observations")# The #x^2# is positive so the general graph shape is #uu# Consider the generalised form of #y=ax^2bxc# The #bx# part of the equation shifts the graph left or right You do not have any #bx# type of value in your equation So the graph is central about the yaxis The #c# part of the equation is of value 1 so it lifts the vertex up from y=0 to y=11 (x) c 2 y 2 (x) where c 1 and c 2 are arbitrary constants, and y 1 (x) and y 2 (x) are two linearly independent solutions (In other words, y 1 and y 2 form a basis of the solution on the interval I ) (4) A particular solution of the differential equation on I is obtained if we assign specific values to c 1 and c 2 in the general solution 2 ndOrder ODE 10 Example Verify that y 1 2x
Math Scene Equations Iii Lesson 3 Quadratic Equations
