The general solution of the second derivative of differential equation y - the first derivative of 4Y + 4Y = e ^ 2x + X

The general solution of the second derivative of differential equation y - the first derivative of 4Y + 4Y = e ^ 2x + X

The homogeneous solution is C1 (x + c2e ^ (2x))

The differential equation y (x + T) = y (x) * y (T), the range of X and t is from negative infinity to positive infinity. When x = 0, the derivative is a, the answer to y (x) is e ^ (AX), and the detailed solution is obtained,
When I calculate y (NX) = [y (x)] ^ n, I have no idea
Detailed explanation, do not guess, I can guess
Add 30 points, thank you!

First of all, let x = t = 0, y (0) = 0 or Y (0) = 1. If y (0) = 0, then let t = 0 know that y (x) is equal to 0, and the derivative will not be a, so y (0) = 1. Then a = Lim [y (T) - Y (0)] / T, when t tends to 0, then y '(x) = Lim [y (x + T) - Y (x)] / T = L

Differential equation y = C1E ^ x + c2e ^ (- 2x) - 1 / 2x-1 / 2 to find y 'is to find the derivative of Y. how to find it? It needs a process

y' = C1e^x-2C2e^(-2x)-1/2

Y '+ 2XY = 4x differential equation, do not set formula or separate variables by finding the homogeneous general solution first and setting C = function back to the original formula
RT

For reference

The general solution of the differential equation y ″ + (Y ′) 2 = 0 is______ ．

Y ″ + (Y ′) 2 = 0, substituting y ″ = dy ′ DX, and shifting the term, we can get: dy ′ DX = − y ′ 2, that is: − dy ′ y ′ 2 = DX, finding the indefinite integral on the left and right sides of the above formula, we can get: 1y ′ = x + C1, C1 is any constant, because: y ′ = dydx, so we can substitute 1y ′ = DXDY into the above formula, that is: DXDY = x + C1, that is

A high number (differential equation) problem!
The differential equation dy / DX + P (x) y = f (x) is known
There are two special solutions: Y1 = - 1 / 4x ^ 2, y2 = - 1 / 4x ^ 2-4 / (x ^ 2)
Find the satisfied P (x), f (x), and give the general solution of the equation
key:
2/x
-x
C4/(x^2)-1/4x^2

(Y1) '= (1 / 4x ^ 2)' = 1 / 2 * x (Y2) '= - 1 / 2 * x + 8 / (x ^ 3) substituting Y1, Y2 and (Y1)' (Y2) 'into the differential equation, we get - 1 / 2 * X-1 / 4 * x ^ 2 * P (x) = f (x) (1) - 1 / 2 * x + 8 / (x ^ 3) - 1 / 4x ^ 2 P (x) + 4 / (x ^ 2) P (x) = f (x) (2) subtracting two expressions, we get 4 / x ^ 2 * P (x) = 8 / x ^ 3, then p (x

Differential equation of higher number problem
Find the general solution of the differential equation dy △ DX + 2XY = 4x,

You're right
But it's easier to understand by separating variables
dy/dx=2x(2-y)
dy/(2-y)=2xdx
The integral of both sides is as follows:
-ln|2-y|=x^2+c1
y=2+ce^(-x^2)

We know that the result of (dy / y) = (DX / x) is often encountered in the solution of differential equation, and then the integral appears ln | y | = ln | x | + C1, which is further transformed into | y | = e ^ C1 | x | and finally y = CX (C = ± e ^ C1). However, the solutions of many exercises are not like this, and I feel that those solutions do not need to consider the absolute value. Why?
For example:
The general solution of the differential equation y '' + (y ') ^ 2 = 0 is__________ .
y=ln(x+C1)+C2.
And I calculate the write as: y = ln|x + c1| + C2. Because I consider the absolute value problem, but the answer does not have an absolute value, very confused. Including a lot of exercises, the solution directly skip the absolute value, this does not write directly to get the following process, I am puzzled
First of all, thank you for your reply. Is the domain of y = ln (x + C1) + C2 different from that of y = ln | x + C1 | + C2? That's why I asked this question

If you think about whether to add absolute value or not, it's not a big problem, or it's unnecessary. As long as X can take all real numbers, y can also take all real numbers. The main reason is that there is an arbitrary constant to play a regulating role! It can play a balancing role! If you ask your first question | y | = e ^ C1 | x | if y has no absolute value, In fact, when you write the answer, y = ln (x + C1) + C2. It's already the default (x + C1) > 0. It doesn't mean much if you add the absolute value or not! You may feel a little vague and don't make it clear, just for reference!
It really can't. when you do the questions in the future, you'd better follow the current method, and then when you write the answers, see if you can remove the absolute value
(just judge if the absolute value is removed, whether the value range of X and y has changed? If there is no change, you can go. The judgment is very simple, just look at their range.)

To solve the ordinary differential equation y + 2XY '+ (x ^ 2) y' '= 0, sit and wait

Let t (T) + T '(y) = t (T) + T' (y)) = t (T) + (T) + (y) = t '(y) = t' (T) + (T) + (T) + (y) = t '(t) + (y) = t' (y) = t '(T) + (T) + (T) + (y) = t' (T) + (T) + (y) = t '(T) + (y) = t' (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (T) + (y)) (T) + (y) = t '(T) + (t') (T) + (t ') (t') (T) + (T) + (T) + (T) + (y ') (t') (T) + (t ') (T) + (T) + (t) + (T) + (T) + (y') (T) + (T) + (T) + (y) = t ')

Ordinary differential equation y = 2XY '+ x ^ 2 / 2 + (y') ^ 2

Input in Matlab: dsolve ('y = 2 * x * dy + x ^ 2 / 2 + dy ^ 2 ','x')
The results are as follows
ans =
-x^2/2
C^2 + C*x - x^2/4