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FODE_16-24

# FODE_16-24 - Tutorial FODE Ordinary First-Order...

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Tutorial FODE: Ordinary First-Order Differential Equations y ( x ) = 1 2 ¡ x 2 ¡ 1 ¢ e x 2 + C e x 2 = 1 2 ¡ x 2 ¡ 1 ¢ + e ¡ x 2 C: y (0) =1 = ) C = 3 = 2 = ) y ( x ) = 1 2 ³ x 2 ¡ 1 +3 e ¡ x 2 ´ : (In the integral for v ( x ) , make the change of variable z = x 2 ; and integrate by parts. c. We must first divide the equation by x to put it in the form of Eqn. 9. Then, q ( x ) = 2 =x and r ( x ) = cos x ¡ 2sin x=x; and u ( x ) = e R x (2 =x 0 ) dx 0 = e 2 ln x = x 2 ; v ( x ) = Z x ¡ x 0 2 cos x 0 ¡ 2 x 0 sin x 0 ¢ dx 0 = x 2 sin x +4 x cos x ¡ 4sin x + C = ) y ( x ) = x 2 sin x +4 x cos x ¡ 4sin x + C x 2 = C x 2 +sin x + 4 x cos x ¡ 4 x 2 sin x: y ( ¼ )= 0 = ) C =4 ¼ = ) y ( x )= 4 ¼ x 2 + sin x + 4 x cos x ¡ 4 x 2 sin x: d. q ( x )= ¡ 1 =x; r ( x )= 2 x 2 = ) u ( x ) = e R x ( ¡ 1 =x 0 ) dx 0 = e ¡ ln x = x ¡ 1 ; v ( x ) = Z x 2 x 0 2 =x 0 dx 0 = x 2 + C = ) y ( x ) = x 2 + C x ¡ 1 = x 3 + Cx: y (1) =2 = ) C = 1 = ) y ( x ) = x 3 + x: Exercise 16. Under these transformations,

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FODE_16-24 - Tutorial FODE Ordinary First-Order...

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