Reduction of order (2nd order linear ODE homogeneous ODE)

[kasse]

Reduce to the 2nd order and solve the eq.

(1-x^2)y'' - 2xy' + 2y = 0 (we know that y(1) = 0)



I don't know what to do here, except that I'm supposed to find y(2). I know the formulas

y(2) = y(1) Int(U) dx

where

U = (e^(-Int(p(x)dx)) / (y(1))^2

Can I use this?

 

[kasse]

Updated

 

[rock.freak667]

To reduce the order, I believe one soltion y_1 must be known and then the other solution is y_2 =vy_1 do you know the at least one solution?

 

[kasse]

To reduce the order, I believe one soltion y_1 must be known and then the other solution is y_2 =vy_1 do you know the at least one solution?

y_1 = x

That's what I meant by Y(1)=0. I don't know how to write these math typings.

 

[rock.freak667]

If y(1)=0 then y(2) = y(1) Int(U) dx then wouldn't y(2) also become 0?

 

[kasse]

Haha, my mistake. y(1)=x, not 0!

 

[rock.freak667]

then it becomes simpler then y_1=x then y_2=vx
then y=vx => \frac{dy}{dx}=x\frac{dv}{dx} + v
and \frac{d^2y}{dx^2}=2\frac{dv}{dx} +x\frac{d^2v}{dx^2}
sub those into the equation and you'll get(x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0

then let any variable you want be \frac{dv}{dx}...say w=\frac{dv}{dx} and then \frac{dw}{dx}=\frac{d^2v}{dx^2} and then you solve from there

i.e (x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0 becomes

(x-x^3)\frac{dw}{dx} + (2-2x)w=0

and what you did there was reduce the 2nd order linear ODE homogeneous eq'n to a first order diff eq'n

 

[kasse]

then it becomes simpler then y_1=x then y_2=vx
then y=vx => \frac{dy}{dx}=x\frac{dv}{dx} + v
and \frac{d^2y}{dx^2}=2\frac{dv}{dx} +x\frac{d^2v}{dx^2}
sub those into the equation and you'll get(x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0

then let any variable you want be \frac{dv}{dx}...say w=\frac{dv}{dx} and then \frac{dw}{dx}=\frac{d^2v}{dx^2} and then you solve from there

i.e (x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0 becomes

(x-x^3)\frac{dw}{dx} + (2-2x)w=0

and what you did there was reduce the 2nd order linear ODE homogeneous eq'n to a first order diff eq'n

I don't understand why I'm supposed to get: (x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0

What I get is: (x-x^3)\frac{d^2v}{dx^2} + (2-4x^2)\frac{dv}{dx} - 4x*v=0

 

[rock.freak667]

ah sorry i forgot to add a term

But this is how the algebra is supposed to be

(1-x^2)(2\frac{dv}{dx}+x\frac{d^2v}{dx^2})-2x(\frac{dv}{dx}+v)+2(vx)=0

2\frac{dv}{dx}-2x^2\frac{dv}{dx}+x\frac{d^2v}{dx^2}-x^3\frac{d^2v}{dx^2}-2x^2\frac{dv}{dx}-2vx+2vx=0

finally giving:

(x-x^3)\frac{d^2v}{dx^2}+(2-4x)\frac{dv}{dx} =0

then sub w=\frac{dv}{dx}

 

[kasse]

OK, I guess you mean

(x-x^3)\frac{d^2v}{dx^2}+(2-4x^2)\frac{dv}{dx} =0

Now then, how do I fine the 2nd solution?

 

[rock.freak667]

Let w= \frac{dv}{dx}
\frac{dw}{dx} = \frac{d^2v}{dx^2}

then sub those in your equation and you will get a variables are separable type of diff eq'n

(x-x^3)\frac{dw}{dx} +(2-4x)w = 0(x-x^3)\frac{dw}{dx}= -(2-4x)w

\frac{1}{w}\frac{dw}{dx} = \frac{-(2-4x)}{(x-x^3)}

integrate both sides w.r.t x now