# A solved 2nd order ODE that i don't understand its solution

1. Jan 22, 2016

### omar yahia

{ i feel that this is Not a smart question and that it is about the basics of something , but i tried to find the that "something" to know about it myself but i couldn't , as i couldn't name the issue , so i couldn't know what to search about
that is why i'm asking here in a forum , so please don't be annoyed if it appeared to be a naive question , thank you :) }
i don't understand why did we change the formula so that only one variable is in it
would it matter if p(x) and r(x) are not constants ? (i.e : something like p(x)= -3/x2 ) after all whether p(x) is a constant or -3/x2 it is after all a function of (x) , then i am on the right formula as the general formula of 2nd order ODE's y" + p(x) y'+ r(x) y= g(x) , right ?
these are 2 solutions (attached ) , mine (the wrong one) , and the other is the model answer (but i see it has an error , m should be 2±j√6 )

Last edited: Jan 22, 2016
2. Jan 22, 2016

### Simon Bridge

... to make the maths easier.

The DE to be solved is: $x^2y'' - 3xy' + 10y = 0$ ... here y is a function of only one variable.
Neither of your worked examples has a p(x) or an r(x) in them, and you don't explain your reasoning as you go.
So it is difficult to see what you are asking. It sounds like you are applying a method to get a solution that you don't understand.
I suspect you are thinking of "reduction of order" ...
http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx

Basically, any nth order DE has a general solution that is a linear combination of n independant solutions ... if you already know one solution to a 2nd order DE, then there is only one other solution to find. You can use the solution you know to help find another solution that is independant of it.
It's much like finding factoring a polynomial by long-division - if you know one of the factors (roots) you can divide it out of the polynomial to get a simpler polynomial to solve.

Last edited: Jan 22, 2016
3. Jan 22, 2016

### omar yahia

indeed , :) thank you a lot ,
one more thing please , this is a quote from Paul's Notes on reduction of order
"Reduction of order requires that a solution already be known. Without this known solution we won’t be able to do reduction of order. "

but the in this problem i wasn't given any solution of y(x) ,
is there any other way to do "reduction of order" that doesn't require already knowing one of the solutions ?

4. Jan 22, 2016

### HallsofIvy

If you have a differential equation such as ay''+ by'+ cy= 0 with constants a, b, and c, then its "characteristic equation" is ar^2+ br+ c= 0. We can get that by arguing that, in order for y, y', and y'' to cancel when multiplied by constants, y, y', and y'' must be the same kind of function. And exponentials have that property. If we try $y= e^{rx}$ then our differential equation becomes $y= e^{rx}(ar^2+ br+ c)= 0$. Since $e^{rx}$ is never 0, we must have $ar^2+ br+ c= 0$, the "characteristic equation above.

But that applies only to "linear equations with constant coefficients". If we have linear equations with non-constant coefficient, as, for example, $x^2y''- 3xy'+ 10y= 0$. That does NOT have constant coefficients so you cannot use the same characteristic equation as above. What you can do is observe that the change of variable $t= ln(x)$ changes the equation from an "equipotential" equation (the power of x multiplying each derivative equals the order of the derivative) in x to a "constant coefficients" equation in t. That is, if $t= ln(x)$, then $dy/dx= (dy/dt)(dt/dx)= (dy/dt)(1/x)$ and $d^2y/dx^2= d(dy/dx)/dx= d((dy/dt)(1/x))/dx= (1/x)(d(dy/dt)/dx)- (1/x^2)(dy/dt)= (1/x^2)(d^2y/dt^2)- (1/x^2)(dy/dt)$ so that $x^2(d^2y/dx2)- 3x(dy/dt)+ 10y= (d^2y/dt^2- dy/dt)- 3(dy/dt)+ 10y= d^2y/dt^2- 4(dy/dt)+ 10y= 0$. The characteristic equation of that "constant coefficients" equation is $r^2-- 4r+ 4= (r- 2)^2= 0$ which has 2 as its only root. That means that the general solution to the constant coefficients solution is $y(t)= Ae^{2t}+ Bte^{2t}$. But $t= ln(x)$ so that $y(x)= Ae^{2ln(x)}+ Bln(x)e^{2ln(x)}= Ae^{ln(x^2)}+ Bln(x)e^{ln(x^2)}= Ax^2+ Bln(x)x^2$ is the general solution to the original equation.

However, linear equations with other non-constant coefficients often cannot be written in terms of "elementary" functions. For example "Bessel's equation", $x^2y''+ xy'+ (x^2- n^2)y= 0$, which is almost an "equipotential" equation, has as its general solution $y(x)= AJ_n(x)+BY_n(x)$ where "$J_n(x)$" and "$Y_n(x)$" are the "Bessel functions of order n of the first and second kind". "Bessel functions" themselves are defined as "the independent solutions to Bessel's equation".

5. Jan 22, 2016

### Simon Bridge

Well... if you had, say: x^2y'' + 2xy' = 0 you could reduce the order without having a solution in mind.

To do the problem you are given, you need to make a guess as to what one of the solutions could be ... test it to make sure it is a solution, and revise your guess accordingly. It takes practise to learn to do this... you need to experience making a lot of wrong guesses before you figure out how to arrive at a good one, maybe even the right one. The various methods for solving DEs you are learning are just ways to make educated guesses. As you see from post #4 the trick is to explore the relationships between the coefficients to get a good guess (I think HallsofIvy has explained the particular trick you were using)... it is not always easy: in fact, it can be arbitrarily difficult.

6. Jan 22, 2016

### omar yahia

thank you very much sir for your time and for this detailed explanation :)

7. Jan 22, 2016

### omar yahia

indeed , as long as i don't have a "master key" method that i can apply to all different kinds of DE's i will have to use guessing sometimes

8. Jan 23, 2016

### omar yahia

ohh ! i have just noticed when i was trying the solution
the characteristic equation should be r^2 -4r + 10 =0 +10 not +4 , right ?