What exactly is a null solution and particular solution?

[goonking]

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

 

[Ssnow]

''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 $\overline{y}=-\frac{1}{2}$ ...

 

[goonking]

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(Ce^4t)

??

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

 

[Ssnow]

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(Ce^4t)

??

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.

 

[Mark44]

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 = Ce^4t. 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.

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.

 

[goonking]

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 = Ce^4t.

I understood everything you said but this.
this means y' - 4y = 0
therefore y' = 4y?

 

[Mark44]

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.