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353_pdfsam_math 54 differential equation solutions odd

# 353_pdfsam_math 54 differential equation solutions odd -...

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Exercises 6.1 Integration yields v 1 ( x ) = w 1 ( x ) dx = ( 3 e 3 x ) dx = e 3 x , v 2 ( x ) = w 2 ( x ) dx = ( 2 e 2 x ) dx = e 2 x , where we have chosen zero integration constants. (d) With functions v 1 ( x ) and v 2 ( x ) obtained in (c), we have y 1 ( x ) = v 1 ( x ) e x = e 3 x e x = e 2 x , y 2 ( x ) = v 2 ( x ) e x = e 2 x e x = e 3 x . To show that the functions e x , e 2 x , and e 3 x are linearly independent on ( −∞ , ), we can use the approach similar to that in Problem 7. Alternatively, since these functions are solutions to the differential equation (32), one can apply Theorem 3, as we did in Problem 15. To this end, W e x , e 2 x , e 3 x ( x ) = e x e 2 x e 3 x e x 2 e 2 x 3 e 3 x e x 4 e 2 x 9 e 3 x = e x e 2 x e 3 x 1 1 1 1 2 3 1 4 9 = 30 e 2 x = 0 on ( −∞ , ) and so the functions e x , e 2 x , and e 3 x are linearly independent on ( −∞ , ). 33. Let y ( x ) = v ( x ) e
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