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Differential Equations Notes

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

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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!

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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.
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