Can Mathematica be used to solve the differential equation: y '' - ay = 0 (a is an unknown number)? And then there are several unknowns in the equation of many variables. I want to find one of them and express it in others,

Can Mathematica be used to solve the differential equation: y '' - ay = 0 (a is an unknown number)? And then there are several unknowns in the equation of many variables. I want to find one of them and express it in others,

&Mathematica can solve your differential equation. I've solved it for you, and I'll show you the results. You can also contact me and discuss other things [email protected]
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How to judge the order of differential equation
 

Second order

Find the order of the differential equation x ^ 3 (y '' ') ^ 4-yy' = 0,

The order of a differential equation is the highest order of the derivative of the unknown variable contained in the equation

How to find the limit of LiMn / N + 1
N tends to infinity

Divided by N at the same time, or the law of Robida, the answer is 1

Given that the point m (2-3m, 1-2m) is in the fourth quadrant, try to determine the value range of M,

Cherry Blossom Yu:
∵ point m (2-3m, 1-2m) is in the fourth quadrant
∴2-3m>0 ①
1-2m

limn→∞(1+2^n+3^n)^(1/n)

Pinch theorem
3^n < 1+2^n+3^n < 3*3^n
3 < (1+2^n+3^n)^(1/n) < 3*3^(1/n)
The limit at both ends is 3
So the original formula = 3

1. If angle α is the second quadrant angle and the end of angle α passes through point P (2m-1,3m-1), find the value range of real number M
2. Write the set of the first quadrants
3. Given Tan α = 2, find the values of the following formulas: (1) sin α + cos α / sin α - cos α; (2) sin α * cos α

(1) If angle α is the second quadrant angle,
2 M-1 ＜ 0, 3 M-1 ＞ 0,
The solution is 1 / 3 < m < 1 / 2
(2) If α is the first quadrant angle,
Then 2m-1 > 0, 3m-1 > 0,
α={m｜m∈(1/2,+∞)}
(3) There are two methods: 1. Using Pythagorean theorem to find the ratio of each side of the triangle with Tan α = 2
2. Tan α = sin α / cos α = 2, that is sin α = 2cos α, substituting (1) to get 3cos α / cos α = 3
(2) This is more complicated than directly calculating sin α, cos α Tan = α sin α = 2 √ 5 / 5 cos α = √ 5 / 5
sinα*cosα =2/5

limn→∞(1+1/3)(1+1/3^2)(1+3^4)… (1+3^2^n)
rt

limn→∞(1+1/3)(1+1/3^2)(1+1/3^4)… (1+1/3^n)
=(3/2)(1-1/3ⁿ)
=3/2

If point a (2m + 4,2m-6) is in the fourth quadrant, find the value range of M

Because a is in the fourth quadrant,
So we can get: 2m + 4 > 0
2m-6＜0
The solution is - 2 < m < 3

Limn tends to infinity (n + 3 / N + 1) ^ n / 2 =?

Which one are you asking? I wrote two according to my understanding