Y''+4y=4/sin^2x y(pi/4)=2 y'(pi/4)=pi

Предыдущий
вопрос Следующий
вопрос

To solve this differential equation, we first need to find the general solution by integrating the given equation.

From the given equation y'' + 4y = 4/sin^2(x), we can rewrite it as:

y'' = 4/sin^2(x) - 4y

Now, let's integrate both sides:

∫ y'' dx = ∫ (4/sin^2(x) - 4y) dx

Integrating the left side, we get:

y' = ∫ (4/sin^2(x) - 4y) dx
y' = 4cot(x) - 4y + C where C is a constant of integration

Next, we integrate again:

y = ∫ (4cot(x) - 4y + C) dx
y = 4ln|sin(x)| - 4y + Cx + D where D is another constant of integration

Now, given the initial conditions y(pi/4) = 2 and y'(pi/4) = pi, we can substitute these values into the general solution to solve for the constants C and D.

When x = pi/4:

y = 4ln|sin(pi/4)| - 4(2) + C(pi/4) + D
y = 4ln(1/sqrt(2)) - 8 + C(pi/4) + D

When x = pi/4:

y' = 4cot(pi/4) - 4y + C
y' = 4 - 8 + C
pi = -4 + C
C = pi + 4

Now we substitute C back into the general solution:

y = 4ln|sin(x)| - 4y + (pi + 4)x + D

Therefore, the solution to the differential equation is:

y = 4ln|sin(x)| - 4y + (pi + 4)x + D