Question

Solve the given initial value problem. y′′+4y′+29y=0; y(0)=3, y′(0)=-5

Answer 1

To solve the given initial value problem y'' + 4y' + 29y = 0, we can use the characteristic equation method. The particular solution to the initial value problem is
`y = e^(-2x)(3cos(5x) - sin(5x))`

Step-by-step explanation:

To solve the given initial value problem y'' + 4y' + 29y = 0, we can use the characteristic equation method. First, we assume a solution of the form y = e^(rt). Substituting this into the differential equation, we get the quadratic equation r^2 + 4r + 29 = 0. Solving this quadratic equation, we find that the roots are complex, r = -2 +/- 5i.

Therefore, the general solution to the differential equation is
`y = e^(-2x)(C1cos(5x) + C2sin(5x))` determined by the initial conditions.

Now, we can use the given initial conditions y(0) = 3 and y'(0) = -5 to find the particular solution. Plugging these values into the general solution, we get the following equations:

3 = C1
-5 = -2C1 + 5C2

Solving these equations simultaneously, we find that C1 = 3 and C2 = -1. Therefore, the particular solution to the initial value problem is
`y = e^(-2x)(3cos(5x) - sin(5x))`