# GS of Differential Equation

1. May 1, 2006

### jamesbob

$$\ddot{x} + 10\dot{x} + 34x = 50e^{-t}$$

Im stuck on getting the homogeneous equation. Im used to equations with just first derivatives, not second derivatives as well. Usually, i'd set the left hand side equal to zero and say:

$$set \left x = Ae^{\lambda{t}} \left and \left so \left \dot{x} = A \lambda e ^{\lambda{t}}$$

id then put in the coefficients and get something like LAMBDA = -2.

But when theres a second derivative i have an extra lamba squared expression and therefore a polynomial that is equal to zero. This results in 2 solutions - is that okay for the homogeneous equation? Il work thru this question as i would:

$$\ddot{x} + 10\dot{x} + 34x = 50e^{-t}$$

$$Homogeneous: \ddot{x} + 10\dot{x} + 34x = 0$$

$$set: \left x = Ae^{\lambda{t}} \left and \left \dot{x} = A \lambda e^{\lambda{t}} \left and \left \ddot{x} = A \lambda^2 e^{\lambda{t}}$$

putting in the coef's we have:

$$A \lambda^2 e^{\lambda{t}} + 10A \lambda e^{\lambda{t}} + 34Ae^{\lambda{t}} = 0$$

this simplifies to:

$$\lambda^2 + 10\lambda + 34 = 0$$

Do i now perform the quadratic formula to get 2 results for LAMBDA thus giving me something along the lines of

$$Homogeneous eq = Ae^{xt} + Ae^{yt}$$ ???

Last edited: May 1, 2006
2. May 1, 2006

### HallsofIvy

Staff Emeritus
You should do that business of setting $y= e^{\lamba}t$ a few times to see how it works. By the time you have done it three or four times, it should have dawned on you that the "characteristic equation" for, say, Ay"+ By'+ C= 0, is Ax2+ Bx+ C= 0 and be able to just write down the equation. Since this is a second order d.e. the characteristic equation is a quadratic equation. Solving that will, in general, give two solutions. That's exactly what you want. The set of all solutions to a second order d.e. form a two-dimensional vector space so you need two independent solutions.

$$Homogeneous eq = Ae^{xt} + Ae^{yt}$$
for several reasons.

For one thing, it is the solution y, not the equation, that is equal to that.
Second, since y was being used as the symbol for the dependent function, it is not a good idea to use y (or x) to represent a number.
Finally, there is no reason to believe that the coefficient of each exponential will be the same (i.e. don't use "A" on both).

A correct solution to the homogeneous equation, assuming the characteristic equation has roots $\lambda_1$ and $\lambda_2$ is
$$y_h(t)= Ae^{\lambda_1 t}+ Be^{\lamba_2 t}$$.

Last edited: May 1, 2006
3. May 2, 2006

### jamesbob

I found a similar question in a past paper that has solutions and from it learned things about the characteristic equation you mentioned, and also to use A + B.

This question is: $$\ddot{x} + 4\dot{x} + 5x = 34e^{2t}$$

You get the characteristic quadratic: $$\lambda^2 + 4\lambda + 5 = 0$$

So $$\lambda = \frac{-4 \pm 2i}{2} = -2 \pm i$$

We had never been given any tutoring on differentials with second derivatives and thus have never had to deal with an answer as such. In this example, the answers gave the complimentary fucntion:

$$x_{CF} = e^{-2t}[A\cos{t} + B\sin{t}]$$

As i said, i have never seen anything like this before - but i can see that with solutions like this, you use the first as the power of e (ie -2) and for i, you say Acost + Bsint - where t has the same coefficient as i.

Applying this to my original question, where the solution i found were $$-5 \pm 3i$$
I get:
$$x_{CF} = e^{-5t}[A\cos{3t} + B\sin{3t}]$$

Is this correct. If so, can someone explain why these rules apply?

4. May 2, 2006

### HallsofIvy

Staff Emeritus
$$e^{a+ bi}= e^a e^{bi}= e^a(cos(b)+ i sin(b))$$

It just the same as eax except that a is complex and so the exponential of the imaginary part become sine and cosine.

Last edited: May 3, 2006
5. May 2, 2006

### jamesbob

Ah rite i see how that works. If i get two solutions for lambda, is my complimentary function:

$$Ae^{\lambda_1} + Be^{\lambda_2}$$

or

$$Ae^{\lambda_1 + \lambda_2} ?$$

6. May 2, 2006

### jamesbob

Another example of this question type that im stuck on is:

$$\ddot{x} + 7\dot{x} + 12x = 24$$

Find the particular solution where $$x(0) = 3, \left \dot{x}(0) = -2$$

My working is:

$$Homogeneous:$$

$$\lambda^2 + 7\lambda + 12 = 0$$

Find soloutions:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 - 4 \times 12}}{2 \times 1} = \frac{-7 \pm 1}{2}$$

This gives solutions $$x = -3, -4$$

So we have $$x_{CF} = Ae^{-3} + Be^{-4} \left (or \left Ae^{-3 + -4} = Ae^{-7} \left) ?$$

$$Particular \left Integral:$$

This is what i did but im not sure its right:

$$set \left x = \alpha, \Rightarrow \dot{x} = 0 \Rightarrow \ddot{x} = 0$$

Subbing in:

$$12\alpha = 24 \Rightarrow \alpha = 2$$

$$General Solution:$$

$$x(t) = Ae^{-3} + Be^{-4} + 2$$
OR
$$x(t) = Ae^{-7} + 2$$

So that's what i get as a general solution. Im pretty sure this is wrong as if i now try to get the particular solution, what do i do with the values given for $$x$$ and $$\dot{x}$$ ?

Last edited: May 2, 2006
7. May 3, 2006

### HallsofIvy

Staff Emeritus
"Complimentary function"? Do you mean the general solution to the homogeneous equation? As I said before, if $\lambda_1$ and $\lambda_2$ are solutions to the characteristic equation, then
$$Ae^{\lambda_1 t}+ Be^{\lambda_2 t}$$
is the solution to the homogeneous equation. (Don't forget the "t"!)

8. May 3, 2006

### HallsofIvy

Staff Emeritus
I always feel a little foolish using the quadratic formula and then discovering that the roots are integers! 12= 3(4) and 3+ 4= 7. the equation factors as (x+ 4)(x+ 3)= 0.

Actually, neither of those is a function! they are constants. Don't forget the variable!

e-3t and e-4t are solutions to the homogeneous equation. e-7t is NOT.

Aren't you learning any of the theory behind this? The set of all solutions to a homogeneous nth order differential equation for an n-dimensional vector space. Any n independent solutions for a basis for that space so any solution can be written as a linear combination of those solutions. A linear combination is something like Af+ Bg where A and B are numbers.

Last edited: May 3, 2006