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# 285hw3 - HW 31 1 Sec.1.6:3 p xy 0 = y 2 xy Set y y = vx y 0...

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HW 3 1 1. Sec.1.6:3. xy 0 = y + 2 p xy: Set v = y x ;y = vx; y 0 = v 0 x + v: Then x ( v 0 x + v ) = vx + 2 p xvx = vx + 2 x p v ) v 0 x + v = v + 2 p v ) v 0 x = 2 p v: This is a separable equation: dv 2 p v = dx ) 1 2 Z dv p v = Z dx x : 1 2 2 p v = ln j x j + C; and v = (ln j x j + c ) 2 : Answer: y = x (ln j x j + C ) 2 : 2. Sec.1.6:21. y 0 = y + y 3 : If we rewrite it in the form y 0 & y = y 3 ; we see that it is a Bernoulli equation with n = 3 . Hence we substitute v = y 1 & 3 = y & 2 ; y = v & 1 = 2 ; y 0 = & 1 2 v & 3 = 2 v 0 ; y 3 = v & 3 = 2 : This gives & 1 2 v & 3 = 2 v 1 & v & 1 = 2 = v & 3 = 2 ; 1 Note to the grader: please grade the following problems: Sec.1.6:49, Sec.2.1:7, Sec. 2.3: 10, Sec. 3.1: 15, Sec. 3.1: 23 (20 pts for each problem) 1

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then multiplication by ( & 2 v 3 = 2 ) produces the linear equation v 1 = 2 v = & 2 with integrating factor & ( x ) = e R 2 dx = e 2 x : Hence e 2 x + 2 e 2 x v = & 2 e 2 x ) D x ( e 2 x v ) = & 2 e 2 x ) e 2 x v = & Z e 2 x d (2 x ) ) e 2 x v = & Z e 2 x d (2 x ) ) e 2 x v = & e 2 x + c ) v = ce & 2 x & 1 ) 1 y 2 = ce & 2 x & 1 : Answer : y 2 = 1 ce & 2 x & 1 3. Sec.1.6:23. xy 0 + 6 y = 3 xy 4 3 ) y 0 + 6 x y = 3 y 4 3 , Bernoulli&s equation, n = 4 3 . Let v = y 1 & 4 3 = y & 1 3 . Then y = v & 3 . Then y 0 = & 3 v 0 v & 4 , and we have y 0 + 6 x y = 3 y 4 3 ) & 3 v 0 v & 4 + 6 x v & 3 = 3 v & 4 ) & 3 v 0 + 6 x v = 3 ) v 0 & 2 x v = & 1 : Integrating factor: & ( x ) = e R & 2 x dx = 1 x 2 . Then 1 x 2 v 0 & 2 x 1 x 2 v = & 1 x 2 ) & v 1 x 2 ± 0 = & 1 x 2 v 1 x 2 = Z & & 1 x 2 ± dx = 1 x + C ) v ( x ) = x + Cx 2 . Then y ( x ) = v & 3 = 1 ( x + 2 ) 3 : 2
4. Sec.1.6:29. Here we assume that x > 0 . Let u = sin y . Then u 0 = cos yy 0 . Hence we have (2 x sin y cos y ) y 0 = 4 x 2 + sin 2 y ) 2 xuu 0 = 4 x 2 + u 2 ) u 0 & 1 2 x u = 2 xu & 1 . This a Bernoulli equation with n = & 1 . Let v = u 1 & n = u 2 . Then u = v 1 2 and u 0 = 1 2 v & 1 2 v 0 . Now u 0 & 1 2 x u = 2 xu & 1 ) 1 2 v & 1 2 v 0 & 1 2 x v 1 2 = 2 xv & 1 2 ) 1 2 v 0 & 1 2 x v = 2 x ) v 0 & 1 x v = 4 x: Thus, we arrive at the following linear equation: v 0 & 1 x v = 4 x: Integrating factor: & ( x ) = e R & dx x = e & ln j x j = 1 x . Next 1 x v 0 & 1 x 2 v = 4 ) & 1 x v ± 0 = 4 ) Z & 1 x v ± 0 dx = 4 Z dx ) 1 x v = 4 x + C ) v ( x ) = 4 x 2 + Cx . Now, because v = u 2 , we have u 2 = 4 x 2 + . Because u = sin y , we have the answer: sin 2 y = 4 x 2 + Cx: 3

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5. Sec.1.6:35. & x 3 + y x ± dx + ² y 2 + ln x ³ dy = 0 : I. @ ² x 3 + y x ³ @y = 1 x and @ ( y 2 + ln x ) @x = 1 x : Hence the equation is exact. II. We have F x = x 3 + y x and F y = y 2 + ln x . First we integrate w.r.t. x : F ( x;y ) = Z & x 3 + y x ± dx = 1 4 x 4 + y ln x + C ( y ) . Now we use F y = y 2 + ln x : ´ 1 4 x 4 + y ln x + C ( y ) µ 0 y = y 2 + ln x; or ln x + C 0 ( y ) = y 2 + ln x ) C 0 ( y ) = y 2 ) C ( y ) = y 3 3 + C . Hence F ( ) = 1 4 x 4 + y ln x + y 3 3 + C . Answer: implicit solution is given by 1 4 x 4 + y ln x + y 3 3 = C .
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285hw3 - HW 31 1 Sec.1.6:3 p xy 0 = y 2 xy Set y y = vx y 0...

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