Y'' + 4y = 4/sin2x y(пи/4)=2 y'(пи/4)=пи

Предыдущий
вопрос Следующий
вопрос

To solve this differential equation, we can start by first finding the general solution by separating the variables:

Y' + 4Y = 4/sin(2x)

Now we determine the integrating factor by solving the characteristic equation, which is given by:

r + 4 = 0
r = -4

So the integrating factor is e^(-4x). Multiplying both sides of the differential equation by the integrating factor, we get:

e^(-4x)Y' + 4e^(-4x)Y = 4e^(-4x)/sin(2x)

The left side can then be written as (e^(-4x)Y)'. Integrating both sides with respect to x, we get:

∫(e^(-4x)Y)' dx = ∫4e^(-4x)/sin(2x) dx
e^(-4x)Y = -2csc(2x) + C

Now we solve for Y by dividing by e^(-4x):

Y = -2csc(2x)e^(4x) + Ce^(4x)

Given that y(π/4) = 2, we substitute x = π/4:

2 = -2csc(π/2)e^(π) + Ce^(π)
2 = -2(0)e^π + Ce^π
2 = 0 + Ce^π
C = 2/e^π

Now we find the derivative of Y to solve for y'(π/4):

Y' = (-2csc(2x)e^(4x) + Ce^(4x))'
Y' = -2(-2cot(2x)csc(2x)e^(4x) + Ce^(4x)) + 4csc(2x)e^(4x)
y'(π/4) = -2(-2cot(π/2)csc(π/2)e^(π) + 2/e^πe^(π)) + 4csc(π/2)e^(π)

Since cot(π/2) = 0, csc(π/2) = 1, and e^π = e^π, we simplify to:

y'(π/4) = 4

Therefore, the solution to the differential equation y'' + 4y = 4/sin(2x) with the initial conditions y(π/4) = 2 and y'(π/4) = π is:

y = -2csc(2x)e^(4x) + (2/e^π)e^(4x)
y'(π/4) = 4