# Second Order Differential Equations.

• Feb 1st 2011, 01:02 AM
rodders
Second Order Differential Equations.
Can someone give me a convincing explanation as to why when solving a second differential equation:
e.g y" + 3y'-4y = 3x+5

That you must ADD the CF to the PI?
- I know it works but i want to know why? I don't want to just demonstrate it either.

It doesn't make sense to me to just say to complete the solution we need to solve y" + 3y'-4y= 0 . Why zero?

The PI clearly works. But also just because we are after some constants doesn't necessarily mean we should put y" + 3y'-4y=0?
• Feb 1st 2011, 01:27 AM
Ackbeet
I'll give you one good reason: a particular integral has no arbitrary constants. Therefore, you cannot play around with the PI to get it to fit initial or boundary conditions. The complementary function does have arbitrary constants. Hence, you get CF + PI to satisfy initial or boundary conditions.
• Feb 1st 2011, 01:52 AM
rodders
I am not sure i buy that. If we are just after arbitrary constants we can pick any function but why choose the CF? and why must the DE=0?
• Feb 1st 2011, 01:56 AM
Ackbeet
You can't pick any ol' function, because you want your result to satisfy the DE. The CF has two properties that make it ideal for satisfying initial conditions: 1. It has two arbitrary constants. 2. It satisfies the homogeneous DE, which means I can change those constants all day long without changing the fact that the CF + PI still satisfies the DE. If you do that with any other function, the result will not satisfy the DE.
• Feb 1st 2011, 03:18 AM
Prove It
For the simple reason that you have to recognise that the RHS is actually $\displaystyle \displaystyle 0 + 3x + 5$, which means to have the most general solution, you need a solution that satisfies the $\displaystyle \displaystyle 0$ as well as the $\displaystyle \displaystyle 3x+5$.
• Feb 1st 2011, 05:07 AM
rodders
BUT why can you do this now but not in the case below?

Because lets look at a quadratic:

x^2 + 5x +4 =2

You can't say the solution consists of x^2+5x +4 =0
and x^2+ 5x+4 =2

So why can you do this now?
• Feb 1st 2011, 05:13 AM
Ackbeet
Because your second-order ODE operator is still linear, though your quadratic equation is not linear. If Ly is the LHS of your ODE, where L represents the differential operator

$\displaystyle L=D^{2} + 3D-4,$

then you can convince yourself that if $\displaystyle y_{1}$ and $\displaystyle y_{2}$ are both solutions to the homogeneous ODE $\displaystyle Ly=0,$ then any linear combination $\displaystyle ay_{1}+by_{2}$ is also a solution to $\displaystyle Ly=0.$

• Feb 1st 2011, 05:19 AM
rodders
Thanks, that is starting to make sense.. Can you give me another example of a linear operator? A simple one?
• Feb 1st 2011, 05:24 AM
rodders
Quote:

Originally Posted by Ackbeet
Because your second-order ODE operator is still linear, though your quadratic equation is not linear. If Ly is the LHS of your ODE, where L represents the differential operator

$\displaystyle L=D^{2} + 3D-4,$

then you can convince yourself that if $\displaystyle y_{1}$ and $\displaystyle y_{2}$ are both solutions to the homogeneous ODE $\displaystyle Ly=0,$ then any linear combination $\displaystyle ay_{1}+by_{2}$ is also a solution to $\displaystyle Ly=0.$

And how does the PI fit into this notation?
• Feb 1st 2011, 05:27 AM
Ackbeet
Sure. Here's a linear operator:

$\displaystyle A=\begin{bmatrix}0&1\\ 1&0\end{bmatrix},$

which has the effect of taking points $\displaystyle (x,y)$ in the $\displaystyle xy$ plane and reversing their coordinates. That is, $\displaystyle (x,y)\overset{A}{\mapsto}(y,x).$ Example:

$\displaystyle \begin{bmatrix}2\\3\end{bmatrix}=\begin{bmatrix}0& 1\\ 1&0\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix} .$

If you want another simple differential operator, try this:

$\displaystyle L=D-1.$

Here's a nonlinear operator:

$\displaystyle L=(1-|y|)D.$

On to your next post: the PI fits into the notation scheme as follows: the particular integral $\displaystyle y_{p}$ satisfies the following equation:

$\displaystyle Ly_{p}=3x+5,$ going back to your original DE.
• Feb 2nd 2011, 01:36 AM
bugatti79
Quote:

Originally Posted by Ackbeet
Because your second-order ODE operator is still linear, though your quadratic equation is not linear. If Ly is the LHS of your ODE, where L represents the differential operator

$\displaystyle L=D^{2} + 3D-4,$

then you can convince yourself that if $\displaystyle y_{1}$ and $\displaystyle y_{2}$ are both solutions to the homogeneous ODE $\displaystyle Ly=0,$ then any linear combination $\displaystyle ay_{1}+by_{2}$ is also a solution to $\displaystyle Ly=0.$