# To determine r you need the indicial equation at x 0

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// To determine r , you need the indicial equation at x 0 = 0 and its roots. Now the indicial equation is r ( r -1) + p 0 r + q 0 = 0 where Thus, the indicial equation is r 2 - 2 r - 8 = 0, with two roots r 1 = 4 and r 2 = -2. Consequently the two linearly independent solutions provided by Theorem 6.3 look like the following: and

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TEST3/MAP2302 Page 2 of 4 ______________________________________________________________________ 3. (10 pts.) Obtain the general solution to the following linear ODE: By letting x = e t , and w ( t ) = y (e t ), so that y ( x ) = w (ln( x )) for x > 0, the ODE above transforms into the following ODE in w ( t ): w ( t ) 4 w ( t ) 2 t . The corresponding homogeneous equ.: w ( t ) 4 w ( t ) 0. The auxiliary equation: m 2 + 4 = ( m + 2 i )( m - 2 i ) = 0 Here’s a fundamental set of solutions for the corresponding homogeneous equation: { cos(2 t ),sin(2 t ) } The driving function of the transformed equation is a U.C. function. By muttering the appropriate incantation and waving your magic writing utensil over the exam, you find that w p ( t ) 1 2 t is a particular integral. Consequently, the general solution to the original ODE, the one involving y , is y ( x ) c 1 cos(2 ln( x )) c 2 sin(2 ln( x )) 1 2 ln( x ) for x > 0.
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