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