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Natural and forced response of a differential equation

marcadams267

New member
Aug 26, 2019
9
Greetings everyone, Im a bit new to differential equations and Im 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, im 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.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
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 $x(t)= (3/14)e^{-4t}- (3/14) cos(3t)+ (2/7) sin(3t)$.