{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

367_pdfsam_math 54 differential equation solutions odd

# 367_pdfsam_math 54 differential equation solutions odd -...

This preview shows page 1. Sign up to view the full content.

Exercises 6.3 By diferentiating y p ( x ), we Fnd y 0 p ( x )= C 1 xe x + C 1 e x + C 3 +2 C 4 x y 0 p ( x C 1 xe x C 1 e x C 4 y 0 p ( x C 1 xe x +3 C 1 e x . Substituting these expressions into the original diferential equation, we obtain y 0 p ( x ) 2 y 0 p ( x ) 5 y 0 p ( x )+6 y p ( x C 1 xe x C 1 e x 2 C 1 xe x 4 C 1 e x 4 C 4 5 C 1 xe x 5 C 1 e x 5 C 3 10 C 4 x +6 C 1 xe x C 2 C 3 x C 4 x 2 = e x + x 2 ⇒− 6 C 1 e x +( 4 C 4 5 C 3 C 2 )+( 10 C 4 C 3 ) x C 4 x 2 = e x + x 2 . Equating coeﬃcients yields 6 C 1 =1 C 1 = 1 6 , 6 C 4 C 4 = 1 6 , 10 C 4 C 3 =0 C 3 = 10 C 4 6 = 10 36 = 5 18 , 4 C 4 5 C 3 C 2 C 2 = 4 C 4 +5 C 3 6 = 4(1 / 6) + 5(5 / 18) 6 = 37 108 . Thus, a general solution to the nonhomogeneous equation is given by y ( x y h ( x )+ y p ( x c 1 e x + c 2 e 3 x + c 3 e 2 x 1 6 xe x + 1 6 x 2 + 5 18 x + 37 108 . 7. In Problem 3, a general solution to the associated homogeneous equation was ±ound to be
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online