{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Differential Equations Notes

Differential Equations Notes - Notes on Second Order Linear...

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

View Full Document Right Arrow Icon
Notes on Second Order Linear Differential Equations Stony Brook University Mathematics Department 1. The general second order homogeneous linear differential equation with constant co- efficients looks like Ay + By + Cy = 0, where y is an unknown function of the variable x , and A , B , and C are constants. If A = 0 this becomes a first order linear equation, which we already know how to solve. So we will consider the case A = 0. We can divide through by A and obtain the equivalent equation y + by + cy = 0 where b = B / A and c = C / A . “Linear with constant coefficients” means that each term in the equation is a constant times y or a derivative of y . “Homogeneous” excludes equations like y + by + cy = f ( x ) which can be solved, in certain important cases, by an extension of the methods we will study here. 2. In order to solve this equation, we guess that there is a solution of the form y = e λ x , where λ is an unknown constant. Why? Because it works! We substitute y = e λ x in our equation. This gives λ 2 e λ x + b λ e λ x + ce λ x = 0. Since e λ x is never zero, we can divide through and get the equation λ 2 + b λ + c = 0. Whenever λ is a solution of this equation, y = e λ x will automatically be a solution of our original differential equation, and if λ is not a solution, then y = e λ x cannot solve the differential equation. So the substitution y = e λ x transforms the differential equation into an algebraic equation!
Background image of page 1

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

View Full Document Right Arrow Icon
2 S ECOND O RDER L INEAR D IFFERENTIAL E QUATIONS Example 1. Consider the differential equation y - y = 0. Plugging in y = e λ x give us the associated equation λ 2 - 1 = 0, which factors as ( λ + 1 )( λ - 1 ) = 0; this equation has λ = 1 and λ = - 1 as solutions. Both y = e x and y = e - x are solutions to the differential equation y - y = 0. (You should check this for yourself!) Example 2. For the differential equation y + y - 2 y = 0, we look for the roots of the associated algebraic equation λ 2 + λ - 2 = 0. Since this factors as ( λ - 1 )( λ + 2 ) = 0, we get both y = e x and y = e - 2 x as solutions to the differential equation. Again, you should check that these are solutions.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}