How Do You Solve the Homogeneous Part of Second Order Differential Equations?

[jamesbob]

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

Im stuck on getting the homogeneous equation. I am 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 there's 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} ?

 

[HallsofIvy]

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

Im stuck on getting the homogeneous equation. I am 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 there's 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} ?

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.

I am not crazy about your writing
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}.

 

[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 function:

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?

 

[HallsofIvy]

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

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

 

[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} ?

 

[jamesbob]

Another example of this question type that I am 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:

For this we read off the characteristic quadratic:

\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 I am 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. I am 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} ?

 

[HallsofIvy]

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} ?

"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"!)

 

[HallsofIvy]

Another example of this question type that I am 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:

For this we read off the characteristic quadratic:

\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

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.

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 I am 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. I am 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} ?

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 n^th 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.