Find the particular solution of the differential equation 𝑑 2𝑥 𝑑𝑡 2 + 16𝑥 = 𝑡 + 𝑒 𝑡 , which satisfies the initial conditions 𝑥(0) = 0, 𝑥′(0) = 1⁄17.

The given differential equation is a second-order linear differential equation with constant coefficients. The associated homogeneous equation is 𝑑^2𝑥/𝑑𝑡^2 + 16𝑥 = 0, which has the characteristic equation 𝑟^2 + 16 = 0. The roots of this equation are 𝑟 = ±4𝑖. Therefore, the general solution of the homogeneous equation is 𝑥ℎ(𝑡) = 𝑐1cos(4𝑡) + 𝑐2sin(4𝑡), where 𝑐1 and 𝑐2 are arbitrary constants.

To find the particular solution, we first find a particular solution of the non-homogeneous equation 𝑑^2𝑥/𝑑𝑡^2 + 16𝑥 = 𝑡. A particular solution of this equation can be assumed as 𝑥𝑝(𝑡) = 𝑎𝑡 + 𝑏, where 𝑎 and 𝑏 are constants to be determined. Substituting 𝑥𝑝(𝑡) into the non-homogeneous equation, we get:

𝑎 = 1/16 and 𝑏 = 0

Therefore, a particular solution of the non-homogeneous equation is 𝑥𝑝(𝑡) = (1/16)𝑡.

Next, we find a particular solution of the non-homogeneous equation 𝑑^2𝑥/𝑑𝑡^2 + 16𝑥 = 𝑒^𝑡. A particular solution of this equation can be assumed as 𝑥𝑝(𝑡) = 𝑎𝑒^𝑡, where 𝑎 is a constant to be determined. Substituting 𝑥𝑝(𝑡) into the non-homogeneous equation, we get:

𝑎 = 1/17

Therefore, a particular solution of the non-homogeneous equation is 𝑥𝑝(𝑡) = (1/17)𝑒^𝑡.

The general solution of the non-homogeneous equation is 𝑥𝑛(𝑡) = 𝑥𝑝(𝑡) + 𝑥ℎ(𝑡) = (1/16)𝑡 + (1/17)𝑒^𝑡 + 𝑐1cos(4𝑡) + 𝑐2sin(4𝑡).

Using the initial conditions 𝑥(0) = 0 and 𝑥′(0) = 1/17, we get:

𝑐1 + 1/17 = 0 and 4𝑐2 + 1/16 = 1/17

Solving these equations, we get 𝑐1 = −1/17 and 𝑐2 = 1/(64π). Therefore, the particular solution of the differential equation 𝑑^2𝑥/𝑑𝑡^2 + 16𝑥 = 𝑡 + 𝑒^𝑡, which satisfies the initial conditions 𝑥(0) = 0, 𝑥′(0) = 1/17 is:

𝑥(𝑡) = (1/16)𝑡 + (1/17)𝑒^𝑡 − (1/17)cos(4𝑡) + (1/(64π))sin(4𝑡).