ДИФФЕРЕНЦИАЛЬНОЕ УРАВНЕНИЕ
y' = y ctgx +2 cosx

Предыдущий
вопрос Следующий
вопрос

This differential equation is a first-order linear ordinary differential equation. To solve it, we can use the method of integrating factors.

First, rearrange the equation to get it into standard form:
y' - y ctgx = 2 cosx

Next, calculate the integrating factor, which is given by the exponential of the integral of the coefficient of y:
μ(x) = e^(∫-ctgx dx) = e^(-ln|sinx|) = 1/sinx

Now, multiply both sides of the equation by the integrating factor and rewrite it in its exact form:
1/sinx * y' - y/sinx ctgx = 2cosx/sinx
d/dx(y/sinx) = 2cosx/sinx

Integrate both sides with respect to x:
y/sinx = 2ln|sinx| + C

Multiply through by sinx to solve for y:
y = 2sinx ln|sinx| + Csinx

Therefore, the general solution to the differential equation y' = y ctgx + 2 cosx is y = 2sinx ln|sinx| + Csinx, where C is an arbitrary constant.