MHB Natural and forced response of a differential equation

[marcadams267]

Greetings everyone, I am a bit new to differential equations and I am trying to solve for the natural and forced response of this equation:

dx/dt+4x=2sin(3t) ; x(0)=0

Now I know that for the natural response I set the right side of the equation equal to 0, so I get
dx/dt+4x=0, thus the characteristic equation is m+4=0 and I get -4 as the root and the solution to the natural response is Ae^-4t and since x(0)=0, A must be 0.
So the natural response is just 0.

However, I am not sure about how to get the forced response, which is when the right side of the equation is not set to 0. Any help is appreciated, thank you.

 

[HOI]

You have correctly found that the general solution to the "associated homogeneous equation" is $x(t)= Ae^{-4t}$. Now you need to find a single solution to the entire equation. The simplest method for getting a solution to the entire equation is "Undetermined Coefficients". Since the "non-homogeneous part" is $2sin(3x)$, "try" a solution of the form $x= Acos(3t)+ Bsin(3t)$. Then $dx/dt= -3Asin(3t)+ 3Bcos(3t)$ and the equation becomes $dx/dt+ 4x= -3Asin(3t)+ 3Bcos(3t)+ 4Acos(3t)+ 4Bsin(3t)= (3B+ 4A)cos(3t)+ (-3A+ 4B)sin(3t)= 2sin(3t)+ 0cos(3t)$ so we must have 3B+ 4A= 0 and -3A+ 4B= 2. Solve those two equations for A and B.

Multiply the first equation by 3: 9B+ 12A= 0. Multiply the second equation by 4: -12A+ 16B= 8. Adding those eliminates A: 7B= 2 so B= 2/7. Then 3B+ 4A= 6/7+ 4A= 0 so 4A= -6/7, A= -3/14. The general solution to the entire equation is $x(t)= Ae^{-4t}- (3/14) cos(3t)+ (2/7) sin(3t)$. NOW apply the "initial condition". x(0)=
A- 3/14= 0 so A= 3/14. The solution is $x(t)= (3/14)e^{-4t}- (3/14) cos(3t)+ (2/7) sin(3t)$.