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**Unformatted text preview: **dy dx + p ( x ) y = q ( x ) multiply by the integral factor f = e R p dx and use d dx ( fy ) = f ( y ± + py ) on the left to get d dx ( fy ) = fq. Problem: Solve dy dx + 2 x (1-x 2 ) y = 4 x,-1 ≤ x ≤ 1 . Solution: Find integral factor, inside integral ﬁrst: Z 2 x (1-x 2 ) dx =-ln(1-x 2 ) = ln ( (1-x 2 )-1 ) (simplify!). ————– So e R 2 x ( 1-x 2 ) dx = e ln ( (1-x 2 )-1 ) = ( 1-x 2 )-1 = 1 1-x 2 = f. Multiply by f = 1 1-x 2 and use Main Property: 1 1-x 2 ± dy dx + 2 x (1-x 2 ) y ¶ = 4 x 1-x 2 , d dx ± y 1-x 2 ¶ = 4 x 1-x 2 . 4 Now integration gives y 1-x 2 = (-2 ln (1-x 2 )) + c, so y = (1-x 2 ) (-ln ((1-x 2 ) 2 ) + c ) . Notice that we can check this instance of the Main Property directly (using the product rule)....

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