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

  • Thread starter kasse
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  • #1
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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?
 
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Answers and Replies

  • #2
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Updated
 
  • #3
rock.freak667
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To reduce the order, I believe one soltion [tex]y_1[/tex] must be known and then the other solution is [tex]y_2 =vy_1[/tex] do you know the at least one solution?
 
  • #4
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To reduce the order, I believe one soltion [tex]y_1[/tex] must be known and then the other solution is [tex]y_2 =vy_1[/tex] do you know the at least one solution?

[tex]y_1[/tex] = x

That's what I meant by Y(1)=0. I don't know how to write these math typings.
 
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  • #5
rock.freak667
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If y(1)=0 then y(2) = y(1) Int(U) dx then wouldn't y(2) also become 0?
 
  • #6
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Haha, my mistake. y(1)=x, not 0!
 
  • #7
rock.freak667
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then it becomes simpler then [tex]y_1=x[/tex] then [tex] y_2=vx[/tex]
then y=vx => [tex]\frac{dy}{dx}=x\frac{dv}{dx} + v[/tex]
and [tex]\frac{d^2y}{dx^2}=2\frac{dv}{dx} +x\frac{d^2v}{dx^2}[/tex]
sub those into the equation and you'll get[tex](x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0[/tex]

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

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

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

and what you did there was reduce the 2nd order linear ODE homogeneous eq'n to a first order diff eq'n
 
  • #8
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then it becomes simpler then [tex]y_1=x[/tex] then [tex] y_2=vx[/tex]
then y=vx => [tex]\frac{dy}{dx}=x\frac{dv}{dx} + v[/tex]
and [tex]\frac{d^2y}{dx^2}=2\frac{dv}{dx} +x\frac{d^2v}{dx^2}[/tex]
sub those into the equation and you'll get[tex](x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0[/tex]

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

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

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

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: [tex](x-x^3)\frac{d^2v}{dx^2} + (2-2x)\frac{dv}{dx}=0[/tex]

What I get is: [tex](x-x^3)\frac{d^2v}{dx^2} + (2-4x^2)\frac{dv}{dx} - 4x*v=0[/tex]
 
  • #9
rock.freak667
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ah sorry i forgot to add a term

But this is how the algebra is supposed to be

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

[tex]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[/tex]

finally giving:

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

then sub [tex]w=\frac{dv}{dx}[/tex]
 
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  • #10
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OK, I guess you mean

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

Now then, how do I fine the 2nd solution?
 
  • #11
rock.freak667
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Let [tex]w= \frac{dv}{dx}[/tex]
[tex]\frac{dw}{dx} = \frac{d^2v}{dx^2}[/tex]

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

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


[tex](x-x^3)\frac{dw}{dx}= -(2-4x)w [/tex]

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

integrate both sides w.r.t x now
 

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