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

2024
Views:

74

Pages:

8
Author:

Javon Neal

Question 1 What is the antiderivative of 𝑥 2 x 2 ? Answer 1 Note that: ∫ 𝑥 𝑛 𝑑 𝑥 = 𝑥 𝑛 + 1 𝑛 + 1 + 𝐶 if 𝑛 ≠ − 1 ∫x n dx= n+1 x n+1 +Cif n  =−1 Answer 2 The antiderivative of a function is the integral of that function. So here we have: ∫ 𝑥 2 𝑑 𝑥 ∫x 2 dx Using the rule that ∫ 𝑥 𝑛 𝑑 𝑥 = 𝑥 𝑛 + 1 𝑛 + 1 + 𝐶 ∫x n dx= n+1 x n+1 +C, 𝑛 ≠ − 1 n  =−1, we see that it equals: 𝑥 2 + 1 2 + 1 + 𝐶 = 𝑥 3 3 + 𝐶 2+1 x 2+1 +C= 3 x 3 +C Answer 3 Write 𝑥 2 x 2 as a function. 𝑓 ( 𝑥 ) = 𝑥 2 f(x)=x 2 The function 𝐹 ( 𝑥 ) F(x) can be found by finding the indefinite integral of the derivative 𝑓 ( 𝑥 ) f(x): 𝐹 ( 𝑥 ) = ∫ 𝑓 ( 𝑥 ) 𝑑 𝑥 F(x)=∫f(x)dx Set up the integral to solve. 𝐹 ( 𝑥 ) = ∫ 𝑥 2 𝑑 𝑥 F(x)=∫x 2 dx By the power rule, the integral of 𝑥 2 x 2 with respect to 𝑥 x is 1 3 𝑥 3 3 1 x 3 : 1 3 𝑥 3 + 𝐶 3 1 x 3 +C The answer is the antiderivative of the function 𝑓 ( 𝑥 ) = 𝑥 2 f(x)=x 2 : 𝐹 ( 𝑥 ) = 1 3 𝑥 3 + 𝐶 F(x)= 3 1 x 3 +C Question 2 If 𝑓 ( 𝑛 ) ( 0 ) = ( 𝑛 + 1 ) ! f (n) (0)=(n+1)! for 𝑛 = 0 , 1 , 2 , 3 … n=0,1,2,3…, find the Maclaurin series for 𝑓 f and its radius of convergence. Answer At the given condition: (See attached handwritten derivation): 𝑓 ( 𝑥 ) = ∑ 𝑛 = 0 ∞ ( 𝑛 + 1 ) ! 𝑛 ! 𝑥 𝑛 = ∑ 𝑛 = 0 ∞ ( 𝑛 + 1 ) 𝑥 𝑛 f(x)= n=0 ∑ ∞ n! (n+1)! x n = n=0 ∑ ∞ (n+1)x n Using the ratio test, the radius of convergence is: 𝑅 = 1 R=1 Question 3 How do you find the antiderivative of cos ⁡ 2 ( 𝑥 ) cos 2 (x)? Answer 1 The secret to figuring out this integral is to use an identity, in this case, the cosine double-angle identity. Since cos ⁡ ( 2 𝑥 ) = cos ⁡ 2 ( 𝑥 ) − sin ⁡ 2 ( 𝑥 ) cos(2x)=cos 2 (x)−sin 2 (x), we can rewrite this using the Pythagorean identity to say that cos ⁡ ( 2 𝑥 ) = 2 cos ⁡ 2 ( 𝑥 ) − 1 cos(2x)=2cos 2 (x)−1. Solving this for cos ⁡ 2 ( 𝑥 ) cos 2 (x) shows us that: cos ⁡ 2 ( 𝑥 ) = 1 + cos ⁡ ( 2 𝑥 ) 2 cos 2 (x)= 2 1+cos(2x) Thus: ∫ cos ⁡ 2 ( 𝑥 ) 𝑑 𝑥 = ∫ 1 2 𝑑 𝑥 + ∫ cos ⁡ ( 2 𝑥 ) 2 𝑑 𝑥 ∫cos 2 (x)dx=∫ 2 1 dx+∫ 2 cos(2x) dx Answer 2 We can’t just integrate cos ⁡ 2 ( 𝑥 ) cos 2 (x) as it is, so we want to change it into another form, which we can easily do using trig identities: cos ⁡ 2 ( 𝑥 ) = 1 + cos ⁡ ( 2 𝑥 ) 2 cos 2 (x)= 2 1+cos(2x) Now we can integrate this as: 𝑥 2 + sin ⁡ ( 2 𝑥 ) 4 + 𝐶 2 x + 4 sin(2x) +C Question 4 Discover the function’s initial partial derivatives. 𝑓 ( 𝑥 , 𝑦 ) = 𝑎 𝑥 + 𝑏 𝑦 𝑐 𝑥 + 𝑑 𝑦 f(x,y)= cx+dy ax+by Answer 1 Consider the next function: 𝑓 ( 𝑥 , 𝑦 ) = 𝑎 𝑥 + 𝑏 𝑦 𝑐 𝑥 + 𝑑 𝑦 f(x,y)= cx+dy ax+by Discover the function’s initial partial derivatives: ∂ 𝑓 ∂ 𝑥 = ( 𝑐 𝑥 + 𝑑 𝑦 ) ( 𝑎 ) − ( 𝑎 𝑥 + 𝑏 𝑦 ) ( 𝑐 ) ( 𝑐 𝑥 + 𝑑 𝑦 ) 2 ∂x ∂f = (cx+dy) 2 (cx+dy)(a)−(ax+by)(c) ∂ 𝑓 ∂ 𝑦 = ( 𝑐 𝑥 + 𝑑 𝑦 ) ( 𝑏 ) − ( 𝑎 𝑥 + 𝑏 𝑦 ) ( 𝑑 ) ( 𝑐 𝑥 + 𝑑 𝑦 ) 2 ∂y ∂f = (cx+dy) 2 (cx+dy)(b)−(ax+by)(d) Question 5 Find all the second partial derivatives: 𝑣 = 𝑥 𝑦 𝑥 − 𝑦 v= x−y xy Answer 1 The second partial derivatives are found by differentiating 𝑣 v with respect to 𝑥 x and 𝑦 y: ∂ 2 𝑣 ∂ 𝑥 2 , ∂ 2 𝑣 ∂ 𝑦 2 , ∂ 2 𝑣 ∂ 𝑥 ∂ 𝑦 ∂x 2 ∂ 2 v , ∂y 2 ∂ 2 v , ∂x∂y ∂ 2 v By applying the quotient rule and differentiating step by step, the solutions are derived as shown.