{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

course materials

# course materials - MA 36600 MIDTERM#2 REVIEW Chapter 3 3.1...

This preview shows pages 1–3. Sign up to view the full content.

MA 36600 MIDTERM #2 REVIEW Chapter 3 § 3.1: Homogeneous Equations with Constant Coefficients. An ordinary differential equation in the form d 2 y dt 2 = G t, y, dy dt is called a second order differential equation . If we have initial conditions in the form y ( t 0 ) = y 0 and dy dt ( t 0 ) = y 0 0 we call this system an initial value problem . Such an equation is called a linear equation if G ( t, y 1 , y 2 ) = g ( t ) - q ( t ) y 1 - p ( t ) y 2 = d 2 y dt 2 + p ( t ) dy dt + q ( t ) y = g ( t ) . Otherwise, we call the differential equation nonlinear . We will consider equations in the form a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = f ( t ) . The functions a ( t ), b ( t ), and c ( t ) are called coefficients . We say that the differential equation is homogeneous if f ( t ) = 0 for all t ; otherwise, it is said to be a nonhomogeneous equation . If we have a constant coefficient homogeneous equation, it is in the form a y 00 + b y 0 + c y = 0. We guess that a solution is y = e rt for some constant r ; this yields the characteristic equation a r 2 + b r + c = 0 . If the characteristic equation a r 2 + b r + c = 0 has two distinct real roots r 1 and r 2 , then the solution to the initial value problem a y 00 + b y 0 + c y = 0 , y ( t 0 ) = y 0 , y 0 ( t 0 ) = y 0 0 ; is the function y ( t ) = c 1 e r 1 t + c 2 e r 2 t in terms of the constants c 1 = y 0 0 - r 2 y 0 r 1 - r 2 e - r 1 t 0 and c 2 = y 0 0 - r 1 y 0 r 2 - r 1 e - r 2 t 0 . § 3.2: Solutions of Linear Homogeneous Equations; the Wronskian. Consider the initial value problem a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = f ( t ); y ( t 0 ) = y 0 , y 0 ( t 0 ) = y 0 0 . Say that we have an interval I = t R α < t < β such that i. f ( t ) is continuous on I , ii. both b ( t ) and c ( t ) are continuous on I , iii. a ( t ) is continuous yet a ( t ) 6 = 0 on I , and iv. t 0 I . Then there exists a unique solution y = y ( t ) to the initial value problem. This is the Existence and Uniqueness Theorem for linear second order differential equations. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 MA 36600 MIDTERM #2 REVIEW Say that y 1 = y 1 ( t ) and y 2 = y 2 ( t ) are solutions to the homogeneous equation a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = 0 . The Principle of Superposition states that the linear combination y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) is also a solution for any constants c 1 and c 2 . We say that { y 1 , y 2 } is a Fundamental Set of Solutions for the differential equation if this is the general solution. Let y 1 = y 1 ( t ) and y 2 = y 2 ( t ) be any two functions. They are said to be linearly independent if the equation “ c 1 y 1 ( t ) + c 2 y 2 ( t ) = 0 for all t ” has the unique solution c 1 = c 2 = 0. Otherwise, we say y 1 and y 2 are linearly dependent . Let y 1 = y 1 ( t ) and y 2 = y 2 ( t ) be any two differentiable functions. The Wronskian of y 1 and y 2 is W ( y 1 , y 2 ) ( t ) = y 1 ( t ) y 0 2 ( t ) - y 0 1 ( t ) y 2 ( t ) = det y 1 ( t ) y 2 ( t ) y 0 1 ( t ) y 0 2 ( t ) . Say that y 1 = y 1 ( t ) and y 2 = y 2 ( t ) are solutions to the homogeneous equation a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = 0 . Abel’s Theorem states that their Wronskian is W ( y 1 , y 2 ) ( t ) = C exp - Z t b ( τ ) a ( τ ) for some constant C . This implies that the following statements are equivalent: i. y 1 ( t ) and y 2 ( t ) are linearly independent.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

course materials - MA 36600 MIDTERM#2 REVIEW Chapter 3 3.1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online