Second Order Differential Equation (homogenous)

[sandy.bridge]

Homework Statement

Given $y_1(x)=x$ is a solution to $(2x-1)y''-4xy'+4y=0$, find $y(2)$ given $(y(1),y'(1))=(0, 0)$. Utilize method of reduction of order.

I need help with this as I end up getting some ugly (in my mind, anyways) integrals. Thanks in advance!

The Attempt at a Solution

Let $y=y_1v=xv$, hence $y'=v'x+v$, and $y''=2v'+v''x$.

$$4xv'-2v'+2x^2v''-xv''-4xv-4x^2v'+4xv=0$$
$$4xv'-2v'+2x^2v''-xv''-4x^2v'=0$$
Let $w=v'$
After separating I get:
$$\frac{dw}{w}=\frac{4x^2-4x+2}{2x^2-x}dx$$

$$\frac{4x^2-4x+2}{2x^2-x}=2+\frac{A}{x}+\frac{B}{2x-1}$$
where $A(2x-1)+Bx=-2x+2$, hence $(A, B)=(-2, 2)$
Therefore,
$$\frac{dw}{w}=(2-2\frac{1}{x}+2\frac{1}{2x-1})dx$$
$$ln|w|=2x-2ln|x|+2ln|2x-1|+C_1$$
$$w=\frac{e^{2x+C_1}(4x^2-4x+1)}{x^2}$$
Next, I know I am supposed to integrate this to determine y', however, I am unsure of how to go about doing this.

 

[haruspex]

You should be able to determine C₁ at this stage from the initial conditions. The answer is rather interesting, but I'm not sure what the implications are.

 

[sandy.bridge]

Well, $w=v'=\frac{y'-v}{x}=\frac{y'-y/y_1}{x}$, however, the initial values causes this to fall to zero, so I am a little confused.

 

[haruspex]

Well, $w=v'=\frac{y'-v}{x}=\frac{y'-y/y_1}{x}$, however, the initial values causes this to fall to zero, so I am a little confused.

Exactly. A degenerate solution, perhaps?

 

[sandy.bridge]

My apologies. This is the first time I have seen this, and my textbook does not seem to mention anything regarding degenerate solutions. What exactly does this imply? Does that mean I assume $v'=0$ and start again?

 

[haruspex]

It implies to me that with these initial conditions y will be always 0. That's certainly a solution.

 

[sandy.bridge]

Okay, let's say that y(1)=1 instead.
That results in $-1=e^{2x+C_1}$ which is impossible, however, the example that I am looking at states that if y(1)=1 there is a solution. What sort of trick does one do after getting to this point, though? If it were zero I can see how that would indicate that v was a constant since v'=0. However, being that it is equal to -1 with y(1)=1, y'(1)=0, I don't quite see what that implies.

 

[haruspex]

Okay, let's say that y(1)=1 instead.
That results in $-1=e^{2x+C_1}$

No, you correctly had ln|w| on the LHS, not ln(w).

 

[sandy.bridge]

Ah yes, don't know how that got away from me. Which brings me to $C_1=-2$. I still have to do a nasty integral to determine $v$ do I not?

 

[haruspex]

I believe that integral can be reduced/equated to integrating a double exponential. There's nothing special about the range, so it cannot be done analytically.
But how come you changed the b.c. from y(1)=0 to y(1)=1? Which is actually wanted?

 

[sandy.bridge]

There was both that I wanted to go over. Thanks for your help! Now that I know I cannot analytically solve that, I am done.

 

[sandy.bridge]

I have been notified that there is an error as this problem is indeed doable analytically. Here is my adjusted work, however, still seems rather tricky to solve:
Let $y=y_1v=xv$, hence $y'=v'x+v$, and $y''=2v'+v''x$.

$$4xv'-2v'+2x^2v''-xv''-4xv-4x^2v'+4xv=0$$
$$4xv'-2v'+2x^2v''-xv''-4x^2v'=0$$
Let $w=v'$
After separating I get:
$$\frac{dw}{w}=\frac{4x^2-4x+2}{2x^2-x}dx$$

$$\frac{4x^2-4x+2}{2x^2-x}=2+\frac{A}{x}+\frac{B}{2x-1}$$
where $A(2x-1)+Bx=-2x+2$, hence $(A, B)=(-2, 2)$
Therefore,
$$\frac{dw}{w}=(2-2\frac{1}{x}+2\frac{1}{2x-1})dx$$
$$ln|w|+c_1=2x-2ln|x|+ln|2x-1|$$
From here we can utilize $y'=v'x+v$, since $w=v'$ we have $y'(1)=w(1)(1)+1$ and $w(1)=-1$.

$$ln|-1|+c_1=2,c_1=2$$
Therefore,
$$v=\int\frac{e^{2x-2}(2x-1)}{x^2}dx$$

From here you can get the right answer by factoring out e^(-2) and applying integration by parts.

 

[haruspex]

$$v=\int\frac{e^{2x-2}(2x-1)}{x^2}dx$$

From here you can get the right answer by factoring out e^(-2) and applying integration by parts.

I got that far before, but it didn't seem like integration by parts would help. What I didn't consider was whether the nasty parts might cancel, so here goes:
$\int_1^2\frac{e^{2x}(2x-1)}{x^2}dx = 2\int_1^2\frac{e^{2x}}{x}dx - \int_1^2\frac{e^{2x}}{x^2}dx = 2\int_1^2\frac{e^{2x}}{x}dx + [\frac{e^{2x}}{x}]_1^2 - 2 \int_1^2\frac{e^{2x}}{x}dx$
Bingo.

 

[sandy.bridge]

$$e^{-2}[2\int\frac{e^{2x}}{x}dx-\int\frac{e^{2x}}{x^2}dx]=e^{-2}[2\int\frac{e^{2x}}{x}dx-\frac{e^{2x}}{x}-2\int\frac{e^2x}{x}dx]=-\frac{e^{2x-2}}{x}+c_2$$
I used $U=e^{2x}, dU=2e^{2x}dx, dV=1/x^2dx, V=-1/x$

From here I determined $c_2=2,$, and $y(2)=-3.38906$ which was the solution.