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Unformatted text preview: 2 nd order, linear, homogeneous, non-constant coefficient ODEs Example: Cauchy-Euler (Euler) O.D.E. 2 2 2 = + + cy dx dy bx dx y d ax here P(x)= 2 ax , Q(x)=bx, R(x)=c and a, b, c = constants This is one of the few cases where an exact solution can be found without using infinite series. Introduce a change of independent variable. Let t = ln x ( or x=e t ). Use the chain rule to change the ODE for y(x) into an ODE for y(t): dt t dy x dt t y d x dx x y d dt t dy x dx dt dt t dy dx x dy ) ( 1 ) ( 1 ) ( ) ( 1 ) ( ) ( 2 2 2 2 2 2 = = = Then, with this substitution, the ODE above reduces to ) ( ) ( ) ( ) ( 2 2 = + + t cy dt t dy a b dt t y d a This constant coefficient homogeneous ODE can be solved easily for y(t) , and then we substitute t = ln x to find the y(x) solution. For applications to be discussed later in the course, the y(x) solution can be solved easily for y(t) as in section 1.2, and then we substitute t = ln x to find the y(x) solution that we want....
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This note was uploaded on 09/16/2011 for the course ME 303 taught by Professor Serhiyyarusevych during the Spring '10 term at Waterloo.
- Spring '10