What exactly is a null solution and particular solution?

I'd recommend spending some time on algebra, specifically equations and their solution methods, before proceeding.
  • #1
goonking
434
3

Homework Statement


Lets say for example, we are given:
dy/dx - 4y = 2 or y' - 4y = 2 , y(0) = 4
=> M= e^(-4t)
e^(-4t) y' - 4e^(-4t)y = 2 e^(-4t)
e^(-4t) y = -1/2 [ e^-4t ] + Cy = -1/2 + Ce^4tWhen t = 0, y = 4 4 = -1/2 + CC = 4.5therefore... y = -1/2 + 4.5e^4t
Now, is -1/2 the null function or particular? what about for 4.5e^4t?

I was imagine a long period of time before the "system" starts, so t = negative infinity, then the exponential function would approach 0, and we would be left with y = -1/2if t = 0, when the system just starts, we were given y = 4now after a very long period of time, t = infinity

y = infinity
What does this mean? the system has no steady state?

Homework Equations

The Attempt at a Solution

 
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  • #2
''null solution'' means the solution of ##y'-4y=0## that is ##y=ce^{4t}##. With ''particular solution'' it is understood as a function ##\overline{y}## that satisfy the general equation ##y'-4y=2## as example in this case ##\overline{y}=-\frac{1}{2}## ...
 
  • #3
so for the particular solution, we just set y' = 0, then we have -4y = 2, y = -1/2

and for the null solution, we need to set the right side to 0, then y' = 4y = 4(Ce4t)

??

is it possible to have a constant as the null solution?
 
  • #4
goonking said:
so for the particular solution, we just set y' = 0, then we have -4y = 2, y = -1/2

and for the null solution, we need to set the right side to 0, then y' = 4y = 4(Ce4t)

??

is it possible to have a constant as the null solution?

Not exactly, in the search of a particular solution you can start to search in this case for constant solutions as ##\overline{y}=k## substituting in the equation you find that ## \overline{y}'-4\overline{y}=0-4k## that must be ##2## so ##k=-\frac{1}{2}##, your particular solution is ##\overline{y}=-\frac{1}{2}##

Yes it is possible to have constant null solutions, as example the null solution of the differential equation ##y'=0## is a constant.
 
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  • #5
goonking said:

Homework Statement


Lets say for example, we are given:
dy/dx - 4y = 2 or y' - 4y = 2 , y(0) = 4
=> M= e^(-4t)
e^(-4t) y' - 4e^(-4t)y = 2 e^(-4t)
e^(-4t) y = -1/2 [ e^-4t ] + C
y = -1/2 + Ce^4t

When t = 0, y = 4
4 = -1/2 + C
C = 4.5

therefore... y = -1/2 + 4.5e^4t

Now, is -1/2 the null function or particular? what about for 4.5e^4t?
Several of your posted questions have been about this topic, so I'm hopeful I can shed some light on it.

With regard to your example, there are two differential equations.
Homogeneous: y' - 4y = 0
Nonhomogenous: y' - 4y = 2
A homogeneous equation has an expression involving y and its derivatives on one side of the equation, and 0 on the other side.
A nonhomogenous equation has an expression involving y and its derivatives on one side of the equation, and a nonzero expression involving the independent variable (x or t or whatever, but not the dependent variable, which here is y).

The null solution (or as it's more commonly called, the complementary solution) is the solution to the homogeneous equation. In this case, it is y = Ce4t. The particular solution is a solution to the nonhomogeneous equation.
The general solution of a nonhomogeneous equation consists of the complementary solution plus the particular solution.
goonking said:
I was imagine a long period of time before the "system" starts, so t = negative infinity, then the exponential function would approach 0, and we would be left with y = -1/2

if t = 0, when the system just starts, we were given y = 4

now after a very long period of time, t = infinity, y = infinityWhat does this mean? the system has no steady state?
Correct, the system has no steady state. Assuming that ##t \ge 0##, y(t) is steadily increasing. That's because of the increasing exponential function in the complementary solution.
 
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  • #6
Mark44 said:
The null solution (or as it's more commonly called, the complementary solution) is the solution to the homogeneous equation. In this case, it is y = Ce4t.

I understood everything you said but this.
this means y' - 4y = 0
therefore y' = 4y?
 
  • #7
goonking said:
I understood everything you said but this.
this means y' - 4y = 0
therefore y' = 4y?
Yes, of course -- these two equations are equivalent. If you are uncertain about this, you're likely to have great difficulties in a class on differential equations.
 

What is a null solution?

A null solution, also known as a trivial solution, is a solution to a system of equations that satisfies all equations with all variables being equal to zero. It is often considered the simplest solution and may not be the most useful in solving real-world problems.

What is a particular solution?

A particular solution is a solution to a system of equations that satisfies all equations with specific values assigned to the variables. It is often used in solving real-world problems where the values of the variables have a specific meaning or significance.

What is the difference between a null solution and a particular solution?

The main difference between a null solution and a particular solution is the values assigned to the variables. A null solution has all variables equal to zero, while a particular solution has specific values assigned to the variables. A null solution is often considered the simplest solution, while a particular solution is more useful in solving real-world problems.

Can a system of equations have both a null solution and a particular solution?

Yes, a system of equations can have both a null solution and a particular solution. In fact, in most cases, a system of equations will have multiple particular solutions that satisfy the equations with different values assigned to the variables. However, there can only be one null solution for a given system of equations.

How do you determine if a solution is a null solution or a particular solution?

To determine if a solution is a null solution or a particular solution, you can plug the values of the solution into the equations and see if they satisfy all equations with all variables being equal to zero (for a null solution) or with specific values assigned to the variables (for a particular solution).

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