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)

a. 74 degrees

step-by-step explanation:

because the lines through the side and the box in the corners shows that these are congruent triangles and since they are mirrored the 37 degrees is also mirrored so the answer is 37 * 2 which is 74.

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y