Final answer:
To find the general solution of the given differential equation y'' + 2y' + y = 200, we first find the general solution of the associated homogeneous equation y'' + 2y' + y = 0. Then, we find the particular solution by using a suitable method. Lastly, we combine the general solution of the homogeneous equation and the particular solution to get the general solution of the original equation.
Step-by-step explanation:
The given equation is a second-order linear homogeneous differential equation with constant coefficients.
To find the general solution, we can first find the associated homogeneous equation by setting the right side of the equation to 0: y'' + 2y' + y = 0.
The characteristic equation is obtained by substituting y = e^(rx) into the homogeneous equation and solving for r. In this case, the characteristic equation is r^2 + 2r + 1 = 0.
Solving the characteristic equation, we get r = -1 (repeated root).
Therefore, the general solution for the homogeneous equation is y_h = (C1 + C2x)e^(-x), where C1 and C2 are constants.
To find the particular solution to the original equation, we can use the method of undetermined coefficients or variation of parameters.
Once the particular solution is found, it can be added to the general solution of the homogeneous equation to obtain the general solution of the original equation.
Since the equation does not specify initial conditions, the general solution is y = (C1 + C2x)e^(-x) + yp, where C1, C2 are constants and yp is the particular solution.