University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

70

Pages:

4
Author:

Javon Neal

Question 1 Find the derivative of 𝑓 ( 𝑢 ) = 9 𝑒 𝑢 + 20 f(u)=9e u +20: Solution: Differentiate 𝑓 ( 𝑢 ) f(u) term by term: The derivative of 𝑒 𝑢 e u is 𝑒 𝑢 e u . The derivative of a constant 20 20 is 0 0. Apply the chain rule to differentiate 9 𝑒 𝑢 9e u : 𝑑 𝑑 𝑢 [ 9 𝑒 𝑢 ] = 9 ⋅ 𝑒 𝑢 . du d [9e u ]=9⋅e u . Combine results: 𝑓 ′ ( 𝑢 ) = 9 𝑒 𝑢 + 0. f ′ (u)=9e u +0. Final Answer: 𝑓 ′ ( 𝑢 ) = 9 𝑒 𝑢 . f ′ (u)=9e u . Question 2 Write the first-order linear differential equation in standard form: Solution: A first-order linear differential equation has the general form: 𝑑 𝑦 𝑑 𝑥 + 𝑝 ( 𝑥 ) 𝑦 = 𝑞 ( 𝑥 ) , dx dy +p(x)y=q(x), where 𝑝 ( 𝑥 ) p(x) and 𝑞 ( 𝑥 ) q(x) are functions of 𝑥 x. Example: 𝑑 𝑦 𝑑 𝑥 + 𝑥 𝑦 = 𝑒 𝑥 dx dy +xy=e x is in standard form. Final Answer: The standard form is: 𝑑 𝑦 𝑑 𝑥 + 𝑝 ( 𝑥 ) 𝑦 = 𝑞 ( 𝑥 ) . dx dy +p(x)y=q(x). Question 3 Find the derivative of 𝑦 = ( 4 𝑡 + 9 ) 1 2 y=(4t+9) 2 1 : Solution: Use the chain rule to differentiate: 𝑑 𝑦 𝑑 𝑡 = 1 2 ( 4 𝑡 + 9 ) − 1 2 ⋅ 𝑑 𝑑 𝑡 ( 4 𝑡 + 9 ) . dt dy = 2 1 (4t+9) − 2 1 ⋅ dt d (4t+9). Differentiate 4 𝑡 + 9 4t+9: 𝑑 𝑑 𝑡 ( 4 𝑡 + 9 ) = 4. dt d (4t+9)=4. Combine results: 𝑑 𝑦 𝑑 𝑡 = 1 2 ( 4 𝑡 + 9 ) − 1 2 ⋅ 4. dt dy = 2 1 (4t+9) − 2 1 ⋅4. Simplify: 𝑑 𝑦 𝑑 𝑡 = 2 ( 4 𝑡 + 9 ) 1 2 . dt dy = (4t+9) 2 1 2 . Final Answer: 𝑑 𝑦 𝑑 𝑡 = 2 ( 4 𝑡 + 9 ) 1 2 . dt dy = (4t+9) 2 1 2 . Question 4 Find the derivative of 𝑘 ( 𝑧 ) = 14 𝑧 + 12 k(z)=14z+12 using the limit definition: Solution: The derivative 𝑓 ′ ( 𝑥 ) f ′ (x) is defined as: 𝑓 ′ ( 𝑥 ) = lim ⁡ ℎ → 0 𝑓 ( 𝑥 + ℎ ) − 𝑓 ( 𝑥 ) ℎ . f ′ (x)= h→0 lim h f(x+h)−f(x) . For 𝑘 ( 𝑧 ) = 14 𝑧 + 12 k(z)=14z+12: 𝑘 ′ ( 𝑧 ) = lim ⁡ ℎ → 0 14 ( 𝑧 + ℎ ) + 12 − ( 14 𝑧 + 12 ) ℎ . k ′ (z)= h→0 lim h 14(z+h)+12−(14z+12) . Simplify the numerator: 𝑘 ′ ( 𝑧 ) = lim ⁡ ℎ → 0 14 𝑧 + 14 ℎ + 12 − 14 𝑧 − 12 ℎ . k ′ (z)= h→0 lim h 14z+14h+12−14z−12 . 𝑘 ′ ( 𝑧 ) = lim ⁡ ℎ → 0 14 ℎ ℎ . k ′ (z)= h→0 lim h 14h . Simplify further: 𝑘 ′ ( 𝑧 ) = 14. k ′ (z)=14. Final Answer: 𝑘 ′ ( 𝑧 ) = 14. k ′ (z)=14. Question 5 Find the derivatives of 𝑟 = 12 𝜃 − 4 𝜃 3 + 1 𝜃 4 r= θ 12 − θ 3 4 + θ 4 1 : Solution: Differentiate term by term using the power rule: If 𝑑 𝑑 𝜃 ( 𝜃 𝑛 ) = 𝑛 𝜃 𝑛 − 1 dθ d (θ n )=nθ n−1 , then: 12 𝜃 = 12 𝜃 − 1 θ 12 =12θ −1 . 4 𝜃 3 = 4 𝜃 − 3 θ 3 4 =4θ −3 . 1 𝜃 4 = 𝜃 − 4 θ 4 1 =θ −4 . Differentiate each term: 𝑑 𝑑 𝜃 ( 12 𝜃 − 1 ) = − 12 𝜃 − 2 . dθ d (12θ −1 )=−12θ −2 . 𝑑 𝑑 𝜃 ( 4 𝜃 − 3 ) = − 12 𝜃 − 4 . dθ d (4θ −3 )=−12θ −4 . 𝑑 𝑑 𝜃 ( 𝜃 − 4 ) = − 4 𝜃 − 5 . dθ d (θ −4 )=−4θ −5 . Combine results: 𝑑 𝑟 𝑑 𝜃 = − 12 𝜃 − 2 + 12 𝜃 − 4 − 4 𝜃 − 5 . dθ dr =−12θ −2 +12θ −4 −4θ −5 . Final Answer: 𝑑 𝑟 𝑑 𝜃 = − 12 𝜃 2 + 12 𝜃 4 − 4 𝜃 5 . dθ dr =− θ 2 12 + θ 4 12 − θ 5 4 .