Giải phương trình vi phân sau
Y"=2y'+y=e^x

You likely mean to write the differential equation,

y'' - 2y' + y = exp(x)

The homogeneous part

y'' - 2y' + y = 0

has characteristic equation

r² - 2r + 1 = (r - 1)² = 0

with a root at r = 1 with multiplicity 2, so the characteristic solution is

y = C₁ exp(x) + C₂ x exp(x)

For the particular solution, we assume an ansatz

y = ax² exp(x)

where a is an unknown constant. Differentiating the ansatz twice:

y' = ax² exp(x) + 2ax exp(x) = (ax² + 2ax) exp(x)

y'' = (ax² + 2ax) exp(x) + (2ax + 2a) exp(x) = (ax² + 4ax + 2a) exp(x)

Substitute y and its derivatives into the DE:

(ax² + 4ax + 2a) exp(x) - 2 (ax² + 2ax) exp(x) + ax² exp(x) = exp(x)

Solve for a :

(ax² + 4ax + 2a) - 2 (ax² + 2ax) + ax² = 1

2a = 1

a = 1/2

Then the particular solution is

y = 1/2 x² exp(x)

The general solution is then

y = C₁ exp(x) + C₂ x exp(x) + 1/2 x² exp(x)

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

what is the original number? you forgot to add it.