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

2024
Views:

341

Pages:

4
Author:

banenetta0l

Question 1 If 𝑓 ( 𝑥 ) + 𝑥 2 [ 𝑓 ( 𝑥 ) ] 5 = 34 f(x)+x 2 [f(x)] 5 =34 and 𝑓 ( 1 ) = 2 f(1)=2, find 𝑓 ′ ( 1 ) f ′ (1). Answer 1 If 𝑓 ( 𝑥 ) + 𝑥 2 [ 𝑓 ( 𝑥 ) ] 5 = 34 f(x)+x 2 [f(x)] 5 =34: Differentiate both sides: 𝑓 ′ ( 𝑥 ) + 2 𝑥 [ 𝑓 ( 𝑥 ) ] 5 + 5 𝑥 2 [ 𝑓 ( 𝑥 ) ] 4 𝑓 ′ ( 𝑥 ) = 0 f ′ (x)+2x[f(x)] 5 +5x 2 [f(x)] 4 f ′ (x)=0 Factor out 𝑓 ′ ( 𝑥 ) f ′ (x): 𝑓 ′ ( 𝑥 ) ( 1 + 5 𝑥 2 [ 𝑓 ( 𝑥 ) ] 4 ) = − 2 𝑥 [ 𝑓 ( 𝑥 ) ] 5 f ′ (x)(1+5x 2 [f(x)] 4 )=−2x[f(x)] 5 Solve for 𝑓 ′ ( 𝑥 ) f ′ (x): 𝑓 ′ ( 𝑥 ) = − 2 𝑥 [ 𝑓 ( 𝑥 ) ] 5 1 + 5 𝑥 2 [ 𝑓 ( 𝑥 ) ] 4 f ′ (x)= 1+5x 2 [f(x)] 4 −2x[f(x)] 5 Substituting 𝑥 = 1 x=1 and 𝑓 ( 1 ) = 2 f(1)=2: 𝑓 ′ ( 1 ) = − 2 ( 1 ) ( 2 ) 5 1 + 5 ( 1 ) 2 ( 2 ) 4 = − 64 1 + 80 = − 64 81 . f ′ (1)= 1+5(1) 2 (2) 4 −2(1)(2) 5 = 1+80 −64 = 81 −64 . So, 𝑓 ′ ( 1 ) = − 64 81 f ′ (1)= 81 −64 . Question 2 Solve the differential equation 𝑡 𝑦 ′ + ( 𝑡 + 1 ) 𝑦 = 𝑡 ty ′ +(t+1)y=t, 𝑦 ( ln ⁡ 2 ) = 1 y(ln2)=1, 𝑡 > 0 t>0. Answer 1 Divide through by 𝑡 t to simplify: 𝑦 ′ + ( 1 + 1 𝑡 ) 𝑦 = 1 y ′ +(1+ t 1 )y=1 The integrating factor is 𝐼 ( 𝑡 ) = 𝑒 ∫ ( 1 + 1 𝑡 ) 𝑑 𝑡 = 𝑡 𝑒 𝑡 I(t)=e ∫(1+ t 1 )dt =te t . Multiply both sides of the equation by 𝑡 𝑒 𝑡 te t : 𝑑 𝑑 𝑡 ( 𝑡 𝑒 𝑡 𝑦 ) = 𝑡 𝑒 𝑡 . dt d (te t y)=te t . Integrate both sides: 𝑡 𝑒 𝑡 𝑦 = ∫ 𝑡 𝑒 𝑡 𝑑 𝑡 . te t y=∫te t dt. Using integration by parts: ∫ 𝑡 𝑒 𝑡 𝑑 𝑡 = 𝑡 𝑒 𝑡 − ∫ 𝑒 𝑡 𝑑 𝑡 = 𝑡 𝑒 𝑡 − 𝑒 𝑡 + 𝐶 . ∫te t dt=te t −∫e t dt=te t −e t +C. So, 𝑡 𝑒 𝑡 𝑦 = 𝑡 𝑒 𝑡 − 𝑒 𝑡 + 𝐶 te t y=te t −e t +C. Simplify for 𝑦 y: 𝑦 = 1 − 1 𝑡 + 𝐶 𝑡 𝑒 𝑡 . y=1− t 1 + te t C . Apply the initial condition 𝑦 ( ln ⁡ 2 ) = 1 y(ln2)=1: 1 = 1 − 1 ln ⁡ 2 + 𝐶 ln ⁡ 2 ⋅ 2 . 1=1− ln2 1 + ln2⋅2 C . Solve for 𝐶 C: 𝐶 = ln ⁡ 2 ⋅ 2. C=ln2⋅2. Final solution: 𝑦 = 1 − 1 𝑡 + ln ⁡ 2 ⋅ 2 𝑡 𝑒 𝑡 . y=1− t 1 + te t ln2⋅2 . Question 3 Explain why or why not determine whether the following statements are true and give an explanation or counterexample. a) The differential equation 𝑦 + 2 𝑦 = 𝑡 y+2y=t. Answer 1 a) The given differential is: 𝑦 ′ + 2 𝑦 = 𝑡 . y ′ +2y=t. The highest order derivative is the first derivative; therefore, it is a first-order differential equation. The equation is linear because 𝑦 y and its derivative appear to the first power, and no products of 𝑦 y and 𝑦 ′ y ′ are present. The equation is not separable, as it cannot be expressed in the form 𝑔 ( 𝑦 ) 𝑑 𝑦 = ℎ ( 𝑡 ) 𝑑 𝑡 g(y)dy=h(t)dt. Therefore, the statement is true. b) The given differential is 𝑦 ′ = 2 𝑡 2 y ′ =2t 2 . Rearranging terms: 𝑑 𝑦 𝑑 𝑡 = 2 𝑡 2 ⇒ 𝑑 𝑦 = 2 𝑡 2 𝑑 𝑡 . dt dy =2t 2 ⇒dy=2t 2 dt. This is first-order because the highest derivative is first-order. It is linear because 𝑦 ′ y ′ and 𝑡 2 t 2 are in separate terms. It is separable, as it can be written as 𝑑 𝑦 = 2 𝑡 2 𝑑 𝑡 dy=2t 2 dt. Thus, the statement is true.