# Can i apply the same technique as I did with real solutions? D.E. 2nd order

1. Feb 20, 2006

### mr_coffee

Hello everyone!
I had a question, I solved a problem very similar to this one but it has 2 real soltions which took the form of:
y (x) = Ae^rx + Be^(rx)

But now i have repeated roots:
y(x) = e^(rx)*(A + Bx);

HEre is the problem:
Find y as a function of x if
x^2 y'' - 11 x y' + 36 y = 0,

y(1) = -2, y'(1) = -3.
y=

Ivey told me i could should change e^(rx) to just x^r in my last problem which worked great!

Now if i have a repeated root case how would I do this? would it be:
y(x) = x^r(A+Bx)?

2. Feb 20, 2006

### HallsofIvy

Staff Emeritus
Nope.

Remember what I said about making the substitution x= et? Then t= ln x so dy/dx= (dy/dt)(dt/dx)= (1/x)dy/dt. Repeating that,
$$\frac{d^2y}{dx^2}= (1/x^2)\frac{d^2y}{dx^2}- (1/x^2)dy/dx$$

$$\frac{d^2y}{dt^2}- 12\frac{dy}{dx}+ 36y= 0$$.
That's a de with constant coefficients that has the same characteristic equation as you got: r2- 12r+ 36= (r- 6)2= 0.

Okay that equation has solutions, as you know well,
$$y(t)= Ae^{6t}+ Bte^{6t}$$. Now put t= ln x:
Of course, $6t= 6 ln x= ln x^6$ so that $e^{6t}= e^{ln x^6}= x^6$.
$$y(t)= Ax^6+ Bln(x)x^6$$.
You see how that t became ln x?

Last edited: Feb 20, 2006
3. Feb 20, 2006

### mr_coffee

Thanks ivey! i'm not sure if i totally fallowed what u wrote but i'm assuming u ment to tell me that:
y = Ax^6 + Bln(x)*x^6 ?
y' = 6Ax^5 +6Bln(x)^x^5 + Bx^5
Well i used that and with the intial conditions:
y(1) = -2
y'(1) = -3
and i found
A and B.
A = -2
B = -1

So i wrote:
y = -2x^6-ln(x)*x^6
but it didn't like it! Some parts of ur latex arn't showing up, i don't know if its my browser or what, i'm on Linux but still using firefox

THanks!

4. Feb 21, 2006

### HallsofIvy

Staff Emeritus
A is -2 but B is NOT -1! Recalculate that. Be careful with your differentiation.