Differential equations particular solution

[faradayscat]

Homework Statement

Particular solution of

y" - y' - 2y = e^(2x)

Homework Equations

None

The Attempt at a Solution

This makes no sense to me, why do I have to use the solution of the form

y(t) = cxe^(2x)

For the problem above, but when I switch the signs and it becomes

y" - y' + 2y = e^(2x) (notice the +2 in front of y)

The solution becomes y(t) = ce^(2x)

I get that it WORKS, but why? Am I seriously supposed to remember hundreds of forms for solutions when I solve these types of DE's or do I solve these with trial and error? It's ridiculously tedious trying several different forms just to realize it doesn't work at the end.

 

[Mark44]

Homework Statement

Particular solution of

y" - y' - 2y = e^(2x)

Homework Equations

None

The Attempt at a Solution

This makes no sense to me, why do I have to use the solution of the form

y(t) = cxe^(2x)

Because y = ke^(2x) is a solution to the homogeneous equation y'' - y' - 2y = 0.
A basis for the solution space of y'' - y' - 2y = 0 is the set {e^(-x}, e^(2x)}. This means that all possible solutions of the homogeneous equation are represented by y = Ae^(-x) + Be^(2x), with A and B arbitrary constants.

For the nonhomogeneous equation, y'' - y' - 2y = e^(2x), there is no possible multiple of e^(2x) that will leave you with e^(2x) on the right side, since e^(2x) is a solution of the homogeneous equation.

What works in cases like this is to look for a solution of the form y = Kxe^(2x).

For the problem above, but when I switch the signs and it becomes

y" - y' + 2y = e^(2x) (notice the +2 in front of y)

The solution becomes y(t) = ce^(2x)

I get that it WORKS, but why?

It works because you now have a different differential equation. e^(2x) is no longer a solution of the homogeneous equation

Am I seriously supposed to remember hundreds of forms for solutions when I solve these types of DE's or do I solve these with trial and error? It's ridiculously tedious trying several different forms just to realize it doesn't work at the end.

I wrote a couple of Insights articles about how to solve nonhomogeneous DEs. The titles are Solving Nonhomogeneous ODEs Using Annihilators. There's a Part 1 and a Part 2.

 

[RUber]

If you were to solve the homogeneous problem:
$y'' - y' - 2y = 0$, you would solve the polynomial equation $z^2 -z -2 =0$ and get $(z-2)(z+1) = 0$ or $z = 2, -1$.
This would lead you to a solution like:
$y(x) = ce^{2x} + de^{-x}$
Now, since your particular solution need to solve:
$y'' - y' - 2y = e^{2x}$
As Mark mentioned, your right hand side is a solution to the homogeneous DE, there is a lot of good theory about null spaces and basis sets to be discussed in other places.
For a less rigorous approach, you can do the following:
Assume that $y = f(x) e^{2x}$
Applying the DE:
$f''(x)e^{2x} +4f'(x) e^{2x} + 4f(x)e^{2x} -2f(x) e^{2x} - f'(x) e^{2x} - 2f(x)e^{2x} = e^{2x}\\ f''(x) + 3f'(x) =1$
The simplest solution to this is of the form $f(x) = cx.$

In the case of $y'' - y' + 2y = e^{2x}$, you might assume the same form:
$y(x) = g(x) e^{2x}$
Applying the DE:
$g''(x)e^{2x} +4g'(x) e^{2x} + 4g(x)e^{2x} -2g(x) e^{2x} - g'(x) e^{2x} + 2g(x)e^{2x} = e^{2x}\\ g''(x) + 3g'(x) +4g(x)=1$
The simplest solution to this is of the form $g(x) = c.$

 

[epenguin]

Preliminary: you have seen something like that in the case of homogeneous lde's? When you have an eigenvalue equation with a double root?

The following is not the way, but it is certainly a way that has a certain rationale to not be too subject to the criticism "something you have to remember".

When I see something like that with an exponential on the RHS I think if I differentiate that I will again get an exponential on the RHS with the same coefficient of x, so I will be able to eliminate between two equations and get a linear homogeneous one, which I have learned to solve. If I do that I get in fact:

y''' -3y'' + 4y = 0

To see how to 'factorise the operator', think of something maybe more familiar, the algebraic equation

x³ - 3x² + 4 = 0

You can see in various ways how to factorise this. E.g. the coefficients are simple numbers, and you see substituting x = -1, that it is a solution. You soon get I think

(x + 1)(x - 2)² is the factorised form of the LHS of this cubic.

So the homogeneous d.e. is

(D + 1)(D - 2)²y = 0

And maybe you have recently or at some time studied and forgotten how to solve homogeneous lde's like that and can remind yourself. This is giving you the general solution of the whole d.e. btw, not just the complementary function or particular integral.

The advantage IMHO is it easier to hit on when you have forgotten what you're supposed to do.

 

[faradayscat]

Because y = ke^(2x) is a solution to the homogeneous equation y'' - y' - 2y = 0.
A basis for the solution space of y'' - y' - 2y = 0 is the set {e^(-x}, e^(2x)}. This means that all possible solutions of the homogeneous equation are represented by y = Ae^(-x) + Be^(2x), with A and B arbitrary constants.

For the nonhomogeneous equation, y'' - y' - 2y = e^(2x), there is no possible multiple of e^(2x) that will leave you with e^(2x) on the right side, since e^(2x) is a solution of the homogeneous equation.

What works in cases like this is to look for a solution of the form y = Kxe^(2x).
It works because you now have a different differential equation. e^(2x) is no longer a solution of the homogeneous equation

I wrote a couple of Insights articles about how to solve nonhomogeneous DEs. The titles are Solving Nonhomogeneous ODEs Using Annihilators. There's a Part 1 and a Part 2.

If you were to solve the homogeneous problem:
$y'' - y' - 2y = 0$, you would solve the polynomial equation $z^2 -z -2 =0$ and get $(z-2)(z+1) = 0$ or $z = 2, -1$.
This would lead you to a solution like:
$y(x) = ce^{2x} + de^{-x}$
Now, since your particular solution need to solve:
$y'' - y' - 2y = e^{2x}$
As Mark mentioned, your right hand side is a solution to the homogeneous DE, there is a lot of good theory about null spaces and basis sets to be discussed in other places.
For a less rigorous approach, you can do the following:
Assume that $y = f(x) e^{2x}$
Applying the DE:
$f''(x)e^{2x} +4f'(x) e^{2x} + 4f(x)e^{2x} -2f(x) e^{2x} - f'(x) e^{2x} - 2f(x)e^{2x} = e^{2x}\\ f''(x) + 3f'(x) =1$
The simplest solution to this is of the form $f(x) = cx.$

In the case of $y'' - y' + 2y = e^{2x}$, you might assume the same form:
$y(x) = g(x) e^{2x}$
Applying the DE:
$g''(x)e^{2x} +4g'(x) e^{2x} + 4g(x)e^{2x} -2g(x) e^{2x} - g'(x) e^{2x} + 2g(x)e^{2x} = e^{2x}\\ g''(x) + 3g'(x) +4g(x)=1$
The simplest solution to this is of the form $g(x) = c.$

Preliminary: you have seen something like that in the case of homogeneous lde's? When you have an eigenvalue equation with a double root?

The following is not the way, but it is certainly a way that has a certain rationale to not be too subject to the criticism "something you have to remember".

When I see something like that with an exponential on the RHS I think if I differentiate that I will again get an exponential on the RHS with the same coefficient of x, so I will be able to eliminate between two equations and get a linear homogeneous one, which I have learned to solve. If I do that I get in fact:

y''' -3y'' + 4y = 0

To see how to 'factorise the operator', think of something maybe more familiar, the algebraic equation

x³ - 3x² + 4 = 0

You can see in various ways how to factorise this. E.g. the coefficients are simple numbers, and you see substituting x = -1, that it is a solution. You soon get I think

(x + 1)(x - 2)² is the factorised form of the LHS of this cubic.

So the homogeneous d.e. is

(D + 1)(D - 2)²y = 0

And maybe you have recently or at some time studied and forgotten how to solve homogeneous lde's like that and can remind yourself. This is giving you the general solution of the whole d.e. btw, not just the complementary function or particular integral.

The advantage IMHO is it easier to hit on when you have forgotten what you're supposed to do.

Thank you so much, I did not realize that e^(2x) was a solution to the homogenous equation.. I probably should have solved the homogeneous case first to check.

 

[LCKurtz]

Thank you so much, I did not realize that e^(2x) was a solution to the homogenous equation.. I probably should have solved the homogeneous case first to check.

More than "probably". That should always be the first step.

 

[faradayscat]

More than "probably". That should always be the first step.

You are right. I have just started solving these types of DE's, thanks for the tips.