Finding a general soution to a differential equation

[Rubik]

How do I find a general solution to a differential equation when the answer is a constant?

 

[JJacquelin]

How do I find a general solution to a differential equation when the answer is a constant?

The answer of what question ? Give an axample.

 

[AlephZero]

y = an arbitrary constant is the general solution to dy/dx = 0

Tell us what equation you are trying to solve, if you want more help.

 

[Rubik]

Something along the lines of y'' - 4y' = 4

 

[JJacquelin]

y'' - 4y' = 4
z = y'
z' - 4z = 4
No difficulty to solve this ODE, thanks to classical method.
z = C* exp(4x) -1
C = constant
Then, the primitives of z(x) :

y = A*exp(4x) - x + B

B = constant ; A = C/4 = constant.

 

[Rubik]

Wow thank you so much.. I was confusing myself and thinking it seemed a little too easy.

Also is this the method of undetermined coefficients?

 

[HallsofIvy]

Yes, it is. Since r= 0 is a solution to the characteristice equation $e^{0x}= 1$, a constant, is already a solution so you, instead of trying a constant to get teh "4" on the right, you multiply by x and try y= Ax. y'= A, y''= 0 so the equation becomes
-4A= 4 and A= -1. That is where the "-x" in the solution came from.

 

[JJacquelin]

Also is this the method of undetermined coefficients?

I suppose that your question is about solving z' - 4z = 4

First, solve the homogeneous equation Z' - 4Z = 0
dZ/dx = 4Z
dZ/Z = 4 dx
ln(Z) = 4x +c
(c = constant)
Z = exp(4x+c)
Z = C*exp(4x)
C = exp(c) = constant

Second, find a particular solution of z' - 4z = 4
z = -1 is obiously a particular solution.

Third, add the pareticular solution to the solutions of the homogeneous equation :
z = Z+(-1)
z = C*exp(4x) -1
This is a well-known way to solve an EDO such as z' - 4z = 4

 

[Rubik]

THANK YOU! This is so helpful :D