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nagle_differential_equations_ISM_Part15

nagle_differential_equations_ISM_Part15 - Chapter 2 32 To...

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Unformatted text preview: Chapter 2 32. To obtain a homogeneous equation, we make a substitution x = u + h , y = v + k with h and k satisfying 2 h + k + 4 = 0 h- 2 k- 2 = 0 ⇒ h =- 6 5 , k =- 8 5 . This substitution yields (2 u + v ) du + ( u- 2 v ) dv = 0 ⇒ dv du = v + 2 u 2 v- u = ( v/u ) + 2 2( v/u )- 1 . We now let z = v/u (so, v = z + uz ) and conclude that z + u dz du = z + 2 2 z- 1 ⇒ u dz du = z + 2 2 z- 1- z =- 2 z 2 + 2 z + 2 2 z- 1 ⇒ Z (2 z- 1) dz z 2- z- 1 =- 2 Z du u ⇒ ln z 2- z- 1 =- 2 ln | u | + C 1 ⇒ ln u 2 z 2- u 2 z- u 2 = C 1 ⇒ ln v 2- uv- u 2 = C 1 ⇒ ln y + 8 5 2- y + 8 5 x + 6 5- x + 6 5 2 = C 1 ⇒ (5 y + 8) 2- (5 y + 8) (5 x + 6)- (5 x + 6) 2 = C, where C = ± 25 e C 1 6 = 0 is any constant. Separating variables, we lost two constant solutions z ( u ), which are the zeros of the polynomial z 2- z- 1. They can be included in the above formula by taking C = 0. Therefore, a general solution is given by (5 y + 8) 2- (5 y + 8) (5 x + 6)- (5 x + 6) 2 = C, where C is an arbitrary constant. 34. In Problem 2, we found that the given equation can be written as a Bernoulli equation, dx dt- 1 2 t x =- t 2 x- 1 . Thus, 2 x dx dt- 1 t x 2 =- t ⇒ ( v = x 2 ) dv dt- 1 t v =- t....
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nagle_differential_equations_ISM_Part15 - Chapter 2 32 To...

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