225_pdfsam_math 54 differential equation solutions odd

225_pdfsam_math 54 differential equation solutions odd - y...

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Exercises 4.6 Therefore, we have y p, 2 ( t )= t 2 +3 . By the superposition principle, we see that a particular solution to the original problem is given by y p ( t )= y p, 1 ( t )+ y p, 2 ( t )=3cos t ln | cos t | +3 t sin t t 2 +3 . Combining this solution with the general solution to the homogeneous equation yields a general solution to the original diFerential equation, y ( t )= c 1 cos t + c 2 sin t t 2 +3+3 t sin t +3cos t ln | cos t | . 17. Multiplying the given equation by 2, we get y 0 +4 y =2tan2 t e t . The nonhomogeneous term, 2 tan 2 t e t , can be written as a linear combination 2 g 1 ( t ) g 2 ( t ), where g 1 ( t )=tan2 t and g 2 ( t )= e t . A particular solution to the equation y 0 +4 y =tan2 t is found in Problem 1, that is,
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Unformatted text preview: y p, 1 ( t ) = − 1 4 cos 2 t ln | sec 2 t + tan 2 t | . A particular solution to y + 4 y = e t can be found using the method of undetermined coefficients. We look for y p, 2 of the form y p, 2 ( t ) = Ae t . Substitution yields ( Ae t ) + 4 ( Ae t ) = e t ⇒ 5 Ae t = e t ⇒ A = 1 5 , and so y p, 2 = (1 / 5) e t . By the superposition principle, a particular solution to the original equation is y p ( t ) = 2 y p, 1 − y p, 2 = − 1 2 cos 2 t ln | sec 2 t + tan 2 t | − 1 5 e t . 221...
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