University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

108

Pages:

2
Author:

banenetta0l

Question 1 Find the differential of each function: (a) 𝑦 = tan ⁡ 𝑡 y=tan t (b) 𝑦 = 1 1 + 𝑥 4 y= 1+x 4 1 Answer (a) 𝑦 = tan ⁡ 𝑡 y=tan t : 𝑑 𝑦 𝑑 𝑡 = sec ⁡ 2 ( 𝑡 ) ⋅ 𝑑 𝑑 𝑡 ( 𝑡 ) = sec ⁡ 2 ( 𝑡 ) ⋅ 1 2 𝑡 = sec ⁡ 2 ( 𝑡 ) 2 𝑡 . dt dy =sec 2 ( t )⋅ dt d ( t )=sec 2 ( t )⋅ 2 t 1 = 2 t sec 2 ( t ) . (b) 𝑦 = 1 1 + 𝑥 4 y= 1+x 4 1 : Using the quotient rule: 𝑑 𝑦 𝑑 𝑥 = 0 − 4 𝑥 3 ( 1 + 𝑥 4 ) 2 = − 4 𝑥 3 ( 1 + 𝑥 4 ) 2 . dx dy = (1+x 4 ) 2 0−4x 3 = (1+x 4 ) 2 −4x 3 . Question 2 Find the general solution of the following differential equation: 𝑑 𝑦 𝑑 𝑥 = ( 2 𝑦 2 + 𝑥 2 𝑒 1 − ( 𝑦 / 𝑥 ) 2 ) ( 2 𝑥 𝑦 ) . dx dy = (2xy) (2y 2 +x 2 e 1−(y/x) 2 ) . Answer Substitute 𝑦 = 𝑣 𝑥 y=vx, 𝑑 𝑦 𝑑 𝑥 = 𝑣 + 𝑥 𝑑 𝑣 𝑑 𝑥 dx dy =v+x dx dv . Replace 𝑦 y and 𝑑 𝑦 𝑑 𝑥 dx dy in the given equation: 𝑣 + 𝑥 𝑑 𝑣 𝑑 𝑥 = 2 ( 𝑣 𝑥 ) 2 + 𝑥 2 𝑒 1 − ( 𝑣 2 ) 2 𝑥 ( 𝑣 𝑥 ) . v+x dx dv = 2x(vx) 2(vx) 2 +x 2 e 1−(v 2 ) . Simplify: 𝑣 + 𝑥 𝑑 𝑣 𝑑 𝑥 = 2 𝑣 2 𝑥 2 + 𝑥 2 𝑒 1 − 𝑣 2 2 𝑣 𝑥 . v+x dx dv = 2vx 2v 2 x 2 +x 2 e 1−v 2 . 𝑣 + 𝑥 𝑑 𝑣 𝑑 𝑥 = 𝑥 2 ( 2 𝑣 2 + 𝑒 1 − 𝑣 2 ) 2 𝑣 𝑥 . v+x dx dv = 2vx x 2 (2v 2 +e 1−v 2 ) . Divide through by 𝑥 2 x 2 : 𝑣 + 𝑥 𝑑 𝑣 𝑑 𝑥 = 2 𝑣 2 + 𝑒 1 − 𝑣 2 2 𝑣 . v+x dx dv = 2v 2v 2 +e 1−v 2 . Subtract 𝑣 v from both sides: 𝑥 𝑑 𝑣 𝑑 𝑥 = 2 𝑣 2 + 𝑒 1 − 𝑣 2 2 𝑣 − 𝑣 . x dx dv = 2v 2v 2 +e 1−v 2 −v. Simplify further: 𝑥 𝑑 𝑣 𝑑 𝑥 = 𝑒 1 − 𝑣 2 − 𝑣 2 2 𝑣 . x dx dv = 2v e 1−v 2 −v 2 . Separate variables and integrate: ∫ 2 𝑣 𝑒 1 − 𝑣 2 − 𝑣 2 𝑑 𝑣 = ∫ 𝑑 𝑥 𝑥 . ∫ e 1−v 2 −v 2 2v dv=∫ x dx . ln ⁡ ∣ 𝑥 ∣ = ∫ 2 𝑣 𝑒 1 − 𝑣 2 − 𝑣 2 𝑑 𝑣 + 𝐶 . ln∣x∣=∫ e 1−v 2 −v 2 2v dv+C. Question 3 Transform the non-autonomous system: 𝑥 1 = 𝑥 , 𝑥 2 = 𝑥 ′ , 𝑥 3 = 𝑥 ′ ′ x 1 =x,x 2 =x ′ ,x 3 =x ′′ 𝑥 ′ ′ ′ = 𝑡 𝑥 . x ′′′ =tx. Answer Let 𝑥 1 = 𝑥 x 1 =x, 𝑥 2 = 𝑥 ′ x 2 =x ′ , 𝑥 3 = 𝑥 ′ ′ x 3 =x ′′ : 𝑥 1 ′ = 𝑥 2 , 𝑥 2 ′ = 𝑥 3 , 𝑥 3 ′ = 𝑡 𝑥 . x 1 ′ =x 2 ,x 2 ′ =x 3 ,x 3 ′ =tx. The system is now a non-autonomous system of first-order differential equations. To convert it to an autonomous system, introduce a new variable 𝑣 = 𝑡 v=t: 𝑣 ′ = 1. v ′ =1. The equations become: 𝑥 1 ′ = 𝑥 2 , 𝑥 2 ′ = 𝑥 3 , 𝑥 3 ′ = 𝑣 𝑥 1 , 𝑣 ′ = 1. x 1 ′ =x 2 ,x 2 ′ =x 3 ,x 3 ′ =vx 1 ,v ′ =1. Now, the system is autonomous with variables ( 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑣 ) (x 1 ,x 2 ,x 3 ,v).