diffeqtest2review - Section 3.4 Reduction of Order Solve...

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Section 3.4 Reduction of Order Solve for the characteristic equation. If the roots are r1,r2 y = C 1 e r1t + C 2 e r2t R1 and r2 complex roots, \ +iu y = C 1 e \ t cos(ut) + C 2 e \ t sin(ut) R1=R2 y = C 1 e r1t + C 2 te r2t Let y = v(t)*(solution already given) Take derivatives y’ and y’’. Plugging into the original equation. Using that t > 0 we use the equation: V’’ = 0 integrating twice we get: V = c1t + c2 Use c1 = 1, c2 = 0 for best C’s possible Plug v(t) into y(t) Second solution is found to be y(t) Section 3.5 Undetermined Coefficients The characteristic polynomial has roots # and # The solutions to the homogenous/complimentary problem are: Y h = c 1 e 3t + C 2 e -t To find a particular solution we try the form: Y(t) = Acos(2t) + Bsin(2t) And take the first and second derivatives Plug into the original equation. Solve for A and B, getting rid of “ghosts”. Once we solve for A and B, we add the particular and the homogenous/complimentary equations. For Equations with a g(x) = -3te -t Try Ate -t , then Ate -t + Bt 2 e -t Section 3.6 Variations of Parameters First we consider the homogenous equation: Y’’ + 4y’ + 4y = 0 Using the characteristic equation:
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R 2 + 4R + 4 = 0 Solve for R. Then its complimentary solution is
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diffeqtest2review - Section 3.4 Reduction of Order Solve...

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