notes 4-3 - of t , sines, cosines, and exponentials. If we...

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SECTION 4.3 METHOD OF UNDETERMINED COEFFICIENTS EXAMPLE. Find the general solution of y 00 - 5 y 0 + 6 y = 60 e - 2 t . Our theory tells us that in order to find the general solution of L [ y ] = g ( t ), we need to (1) solve L [ y ] = 0, and then (2) find any particular specific function y p ( t ) that is a solution of L [ y ] = g ( t ). In our example we know how to do Part (1), so we need a method for producing a y p ( t ). EXAMPLE, CONTINUED. Rewrite the equation using the D -operator, then ask, “What D -operator, if any, will kill the right-hand-side?” Draw appropriate conclusions and finish the problem!
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This method of undetermined coefficients works only for differential equations with constant coefficients for which the g ( t ) on the right side consists of sums of products of powers
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Unformatted text preview: of t , sines, cosines, and exponentials. If we start with L [ y ] = g ( t ), then nd a D-operator K that kills (annihilates!) g ( t ), write down the general solution of KL [ y ] = 0, strike out the solutions of L [ y ] = 0, and determine the remaining coecients by plugging the result into L [ y ] = g ( t ). This works because we can nd a constant-coecient K and because the algebra of constant-coecient operators like K and L is just like polynomials. EXAMPLE. Find the solution of the initial value problem y 00-3 y + 2 y = 8 te 3 t + 20 cos t + 3 t 2 e 2 t , y (0) = 0 , y (0) = 0 HOMEWORK: SECTION 4.3...
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notes 4-3 - of t , sines, cosines, and exponentials. If we...

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