- #1

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y''-5y'+6y=0

Because it can just as easily be written y''-5y'=-6y

I am interested in the meaning of why if we sum y''+(-5y')+6y equals zero. What is the relationship of its second derivative, first derivative, and 6y?

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- Thread starter JaredPM
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- #1

- 20

- 0

y''-5y'+6y=0

Because it can just as easily be written y''-5y'=-6y

I am interested in the meaning of why if we sum y''+(-5y')+6y equals zero. What is the relationship of its second derivative, first derivative, and 6y?

- #2

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y''-5y'+6y=0

Because it can just as easily be written y''-5y'=-6y

I am interested in the meaning of why if we sum y''+(-5y')+6y equals zero. What is the relationship of its second derivative, first derivative, and 6y?

I would say it is similar to solving quadratic equations, saying ax^2+bx+c=0. We could move the c over, but after factoring to (x-x_1)(x-x_2)=0, it is easier to read off the roots. Also, sometimes there is a function of t on the right,

y''-5y'+6y=f(t),

and the business on the left is considered in methods apart and together with f. There's a number of reasons, and at the same time, it's not really a big deal. We can move c over, in ax^2+bx=-c, and we often do if we are completing the square, but it's still usually presented as "=0".

Try reading some of the methods on techniques for solving the

This may give you an idea as to why it's more comfortable to put all the y terms on the left. Also, note that, if anything, it might make more sense to put the y'' on the left, like y''=f(t,y,y'). Compare this with something you may have seen, y'=p(t)y(t)+q(t)=f(t,y).

In retrospect, one might answer your question by saying we write it this way when our differential equation is a "linear" differential equation, meaning, as opposed to any general differential equation which can be written

y^(n)=f(t,y,y'',...,y^(n-1)),

it is of the special form

a_n(t)*y^(n)(t)+a_{n-1}(t)*y^(n-1)(t)+...+a_2(t)*y''(t)+a_1(t)*y'(t)+a_0(t)*y(t)=f(t).

That is, the coefficients are functions of t, and there are no terms like sin(y'') or (y')^3. The form we have written it in is in a form ready to use many of the methods of solving. Why on earth do we have such a bizarre definition of linear? I feel this is because differential equations of the above form, obey a very nice property, in that we can add solutions. In other words, the solution space forms a vector space (offset by a particular solution, so I think we call it affine. Think of it as a plane shifted away from the origin.)

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- #3

HallsofIvy

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So we "try" a solution of the form [itex]y= e^{rx}[/itex] so that [itex]y'= re^{rx}[/itex] and [itex]y''= r^2e^{rx}[/itex]. Putting those into any of the previous forms gives:

[itex]y''- 6y'+ 5y= r^2e^{rx}- 6re^{rx}+ 5e^{rx}= e^{rx}(r^2- 6r+ 5)= 0[/itex] or

[itex]y''+ 5y= e^{rx}(r^2+ 5)= 6y'= 6re^{rx}[/itex] or

[itex]y''= r^2e^{rx}= 6y'- 5y= (6r- 5)e^{rx}[/itex]

In any case, since [itex]e^{rx}[/itex] is not 0 for any x, we can cancel it to get

[itex]r^2- 6r+ 5= 0[/itex] or [itex]r^2+ 5= 6r[/itex] or [itex]r^2= 6r- 5[/itex].

That is a quadratic equation. In the first form, we can factor it to get [itex](r- 5)(r- 1)[/itex] (in itself, perhaps a good reason for writing it in that form). That tells us that r= 1 or r= 5 so that both [itex]e^x[/itex] and [itex]e^{5x}[/itex] satisfy the differential equation.

Now, there is some "logic" to the rest- it can be shown that the set of all solutions to a linear nth order differential equation forms an n dimentional vector space. Here that means that since [itex]e^x[/itex] and [itex]e^{5x}[/itex] are independent solutions,

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