General Form:
The general representation of a 2nd order linear ODE is given by:[ y'' + P(t)y' + Q(t)y = g(t) ]
Homogeneous vs Nonhomogeneous:
If ( g(t) = 0 ), the equation is homogeneous, otherwise it is nonhomogeneous.
Constant Coefficients:
If ( P(t) ) and ( Q(t) ) are constant functions, the equation has constant coefficients; otherwise, it has nonconstant coefficients.
Nonconstant Coefficients, Nonhomogeneous:
( y'' + p(t)y' + q(t)y = g(t) )
Nonconstant Coefficients, Homogeneous:
( y'' + p(t)y' + q(t)y = 0 )
Constant Coefficients, Nonhomogeneous:
( y'' + ay' + by = g(t) )
Constant Coefficients, Homogeneous:
( y'' + ay' + by = 0 )
Remark:
Case IV is typically the easiest and was completely solved in the last lecture.
Case I remains the most general and challenging, having applications in special functions and quantum mechanics.
Notation:
Define differential operator:[ L[y] = y'' + p(t)y' + q(t)y ]
The Wronskian:
Wronskian of two functions ( y_1 ) and ( y_2 ) is given by:[ W[y_1, y_2] = y_1y_2' - y_1'y_2 ]
Properties of Linear Operators:
Linear operators maintain linearity:
( L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)] )
If two functions satisfy the linear equation, their linear combination also satisfies the equation.
Theorem (Superposition):
If ( y_1(t) ) and ( y_2(t) ) are solutions of the ODE ( L[y] = 0 ), then the linear combination ( C_1 y_1(t) + C_2 y_2(t) ) is also a solution for any constants ( C_1 ) and ( C_2 ).
Proof involves demonstrating that the operator ( L ) applied to this linear combination yields zero.
To satisfy initial conditions, the general solution has the form:[ y(t) = C_1 y_1(t) + C_2 y_2(t) ]
Determining Coefficients:
By substituting initial conditions:[ y(t_0) = y_0, , y'(t_0) = y_1 ]
Can solve for constants uniquely if the Wronskian is not zero.
Consider the initial value problem (IVP):[ y'' + P(t)y' + Q(t)y = g(t) ] with continuous functions on interval ( I ) containing ( t_0 ).
Theorems ensure solutions exist and are unique:
The IVP has a solution.
The solution is unique within the interval.
Solutions exist on the largest interval possible where the functions are continuous.
Given IVP:[ (t^2-3)y'' + ty' - (t+3)y = 0 ] with conditions ( y(1) = 2 ) and ( y'(1) = 1 ).
Identify intervals with potential discontinuity where the coefficients are not continuous.
The largest open interval containing ( t_0 ) where the equation is continuously defined is ( (0, 3) ).
Problem Statement: Find two monomial solutions of the equation:[ 2y'' + 3y' - y = 0 ]
Check if they form a fundamental set of solutions and find the general solution.
Guessing a solution of the form ( y(t) = t^n ) leads to the characteristic equation to derive the roots and corresponding functions for the general solution.
Fundamental Set of Solutions:
The solution set provides a means to express general solutions of 2nd order linear ODEs.
Solutions are linear combinations of independent solutions determined by the Wronskian.