Question

Y"+6y'+34=0​

Answer 1

Ответ: y=C1*e^(-3*x)*cos(5*x)+C2*e^(-3*x)*sin(5*x), где C1 и C2 - произвольные постоянные.

Пошаговое объяснение:

Перед нами - однородное ЛДУ 2 порядка с постоянными коэффициентами. Для его решения составляем характеристическое уравнение:

k²+6*k+34=0. Оно имеет комплексные сопряжённые корни k1=-3+5*i и k2=-3-5*i, где i=√(-1). Поэтому общее решение уравнения имеет вид: y=C1*e^(-3*x)*cos(5*x)+C2*e^(-3*x)*sin(5*x).

Answer 2

The differential equation you've given is a second-order linear homogeneous ordinary differential equation with constant coefficients. The equation is:

Y'' + 6Y' + 34 = 0

To solve this differential equation, you can first find the characteristic polynomial and then use it to find the general solution.

The characteristic polynomial for this equation is obtained by replacing Y'' with r^2, Y' with r, and the constant 34 with 0:

r^2 + 6r + 34 = 0

Now, you can use the quadratic formula to solve for the roots of this polynomial:

r = (-6 ± √(6^2 - 4134)) / (2*1) r = (-6 ± √(36 - 136)) / 2 r = (-6 ± √(-100)) / 2 r = (-6 ± 10i) / 2

The roots are complex (with an imaginary part), so you have:

r1 = (-6 + 10i) / 2 = -3 + 5i r2 = (-6 - 10i) / 2 = -3 - 5i

Now, you can write the general solution to the differential equation using these complex roots:

Y(t) = e^(rt) [A cos(ωt) + B sin(ωt)]

Where:

• r1 and r2 are the roots of the characteristic polynomial.
• ω is the imaginary part of the roots (in this case, ω = 5).
• A and B are arbitrary constants that you need to determine based on initial conditions or additional information.

So, the general solution to the differential equation is:

Y(t) = e^(-3t) [A cos(5t) + B sin(5t)]

To find the specific solution for your problem, you would need additional information such as initial conditions or boundary conditions. These conditions would allow you to determine the values of A and B in the general solution.