Another head scratcher

**doke** · November 1st 2006, 02:19 AM

I am trying to set up the appropriate particular solution for this differential equation but don't know where to start. It doesn't even seem like a ode since I only have y terms on the LHS and no derivatives of y.

Any help would be appreciated just to set this up. Thanks

(D-1)^3(D^2-4)y=xe^x +e^2x +e^-2x

**topsquark** · November 1st 2006, 03:57 AM

Originally Posted by doke

I am trying to set up the appropriate particular solution for this differential equation but don't know where to start. It doesn't even seem like a ode since I only have y terms on the LHS and no derivatives of y.

Any help would be appreciated just to set this up. Thanks

(D-1)^3(D^2-4)=xe^x +e^2x +e^-2x

If D is supposed to be a differential operator, we need something like
$(D-1)^3(D^2-4)y$

If not, what is D?

-Dan

**doke** · November 1st 2006, 01:43 PM

Yes, I'm sorry. There is a y at the end of the parenthesis. I just edited my post.

**Soroban** · November 1st 2006, 02:39 PM

Hello, doke!

Are you in self-study?
This notation should have been explained to you in class.

$(D-1)^3(D^2-4)y \:=\:xe^x + e^{2x} +e^{-2x}$

When using the "Method of Operators", they often use this form of notation.

$Dy$ means $\frac{dy}{dx}$ . . . $D^2y$ means $\frac{d^2y}{dx^2}$ . . . $D^3y$ means $\frac{d^3y}{dx^3}$

Example

An equation like $\frac{dy}{dx} - 2y \;= \;x$ can be written: . $Dy - 2y \;=\;x$

Then we can factor out the y (ha!): . $(D - 2)y \;=\;x$

To solve for $y$ , we would "divide by $(D - 2)$ "

A really silly thing to say because $\frac{x}{D - 2}$ has no meaning . . . or does it?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Suppose we have the equation: . $\frac{dy}{dx} - ay \:=\:P(x)$

We can solve it with an integrating factor: . $I \:=\:e^{\int \text{-}a\,dx} \:=\:e^{\text{-}ax}$

Multiply through by $I:\;\;e^{\text{-}ax}\frac{dy}{dx} + ae^{\text{-}ax}y \;=\;e^{\text{-}ax}P(x)$

The left side is the derivative of a product: . $\frac{d}{dx}\left(e^{\text{-}ax}y\right) \;=\;e^{\text{-}ax}P(x)$

Integrate: . $e^{\text{-}ax}y \;=\;\int e^{\text{-}ax}P(x)\,dx$

Therefore: . $\boxed{y \;= \;e^{ax}\int e^{\text{-}ax}P(x)\,dx}$

The same equation can be written: . $Dy - ay \;=\;P(x)$

. . which gives us: . $(D - a)y \;=\;P(x)\quad \Rightarrow \quad y \;=\;\boxed{\frac{P(x)}{D-a}}$

We have now defined that 'rather silly' statement:

. . . . . $\boxed{\boxed{\frac{P(x)}{D - a} \;= \;e^{ax}\int e^{-ax}P(x)\,dx}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to our example . . .

We had: . $(D - 2)y\:=\:x\quad\Rightarrow\quad y \:=\:\frac{x}{D - 2}$

By definition: . $y \;\;= \;\;\frac{x}{D - 2} \;\;= \;\;e^{2x}\!\!\int xe^{-2x}dx$

Integrate by parts and get: . $y \;= \;e^{2x}\left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C\right]$

Therefore: . $y \;= \;-\frac{1}{2}x - \frac{1}{4} + Ce^{2x}\quad\Rightarrow\quad \boxed{y \;= \;-\frac{1}{4}\left(2x + 1 + Ce^{2x}\right)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Now, your problem is inhumanely long and difficult!

You have: . $(D-1)^3(D-2)(D+2)y \:=\:xe^x + e^{2x} +e^{-2x}$

You have: $P(x) \:=\:xe^x + e^{2x} + e^{-2x}$

And you are expected to "divide by $D - 1$ " three times
. . then "divide by $D - 2$ ", then "divide by $D + 2$ ".

. . . I'll wait in the car.

**topsquark** · November 1st 2006, 04:20 PM

$(D-1)^3(D^2-4)y = xe^x + e^{2x} + e^{-2x}$

We need a particular solution to this. Note that:
$(D-1)^3(D^2-4) = D^5 + \alpha D^4 + ... + \beta D + 4$

To get a homogeneous solution for this differential equation, note that all of the coefficients of the D operator are constants. This means that the homogeneous solutions are going to be of the form:
$y_h(x) = Ae^{ax} + Be^{bx} + ...$
where the a, b, ... solve the characteristic polynomial.

As to the solution of the characteristic polynomial, we know them, since the problem gives us a factored version of the operator. The characteristic polynomial is:
$(m - 1)^3(m^2 - 4) = 0$ .
This has roots at the three cube roots of 1 and at m = 2, -2.

This suggests the following form for the particular solution:
$y_p(x) = Cxe^x + Dxe^{2x} + Exe^{-2x}$

We need the first term to produce the $xe^x$ on the RHS of the differential equation. We need the second and third terms the way they are in order to produce the $e^{2x} + e^{-2x}$ . We need to multiply by the extra "x" in this case because both $e^{2x}$ and $e^{-2x}$ are already in the homogeneous solution.

I hope I explained that well enough!

-Dan

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