Answer:
Step-by-step explanation:
(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.
(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.
(C) Example of a second order linear ODE:
M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)
The equation will be homogeneous if K(t)=0 and heterogeneous if 
Example of a second order nonlinear ODE:

(D) Example of a nonlinear fourth order ODE:
![K^4(x) - \beta f [x, k(x)] = 0](https://tex.z-dn.net/?f=K%5E4%28x%29%20-%20%5Cbeta%20f%20%5Bx%2C%20k%28x%29%5D%20%3D%200)
36 divided by 90 gives the percent so its 0.4 which is 40%
Answer:
i don't know
Step-by-step explanation:
Answer:
(a) (x+1)(x-1)
(b)(3x+1)(3x-1)
(c) (x+3)(x+5)
(d)(2x+5)(2x+3)
(e)(x+y)(x-y)
(f) 
Step-by-step explanation:
We have to factorize the following expressions:
(a) x²-1 =(x+1)(x-1) (Answer) {Since we know the formula (a²-b²) =(a+b)(a-b)}
(b) 9x²-1 =(3x+1)(3x-1) (Answer) {Since we know the formula (a²-b²) =(a+b)(a-b)}
(c) x²+8x+15 = x² +3x+5x+15 =(x+3)(x+5) (Answer)
(d) 4x²+16x+15 =4x²+10x+6x+15 = 2x(2x+5) +3(2x+5) =(2x+5)(2x+3) (Answer)
(e) x²-y² =(x+y)(x-y) (Answer)
(f)
(Answer) {Since we know the formula (a²-b²) =(a+b)(a-b)}
Formula for distance between two points is
d∧2=(x2-x1)∧2+(y2-y1)∧2 in our case (x1,y1)=(k,q) and (x2,y2)=(0,0)
d∧2=(0-k)∧2+(0-q)∧2= k∧2+q∧2 => d=√(k∧2+q∧2)
Good luck!!!