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

In summary, to reduce the given second order linear ODE to a first order differential equation, we can use the substitution y_2=vx where v is any variable. This leads to the equation (x-x^3)\frac{d^2v}{dx^2} + (2-4x)\frac{dv}{dx}=0. By substituting w=\frac{dv}{dx} and integrating both sides, we can find the second solution for the given equation.
  • #1
kasse
384
1
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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  • #2
Updated
 
  • #3
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
rock.freak667 said:
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
If y(1)=0 then y(2) = y(1) Int(U) dx then wouldn't y(2) also become 0?
 
  • #6
Haha, my mistake. y(1)=x, not 0!
 
  • #7
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
rock.freak667 said:
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
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
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
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
 

What is the concept of reduction of order in second order linear homogeneous ODE?

Reduction of order is a method used to solve second order linear homogeneous ODEs by reducing the order of the equation to a first order equation. This is done by assuming the solution to be in the form of y = u(x)v(x), where u(x) is a known function and v(x) is an unknown function.

How is the method of reduction of order applied to solve second order linear homogeneous ODEs?

To apply the method of reduction of order, we first find the general solution to the homogeneous equation by using the characteristic equation. Then, we assume the solution to be in the form of y = u(x)v(x) and substitute it into the equation. This reduces the equation to a first order equation in terms of v(x). Solving this equation gives the general solution to the original equation.

What is the advantage of using the reduction of order method to solve second order linear homogeneous ODEs?

The advantage of using this method is that it simplifies the solution process for higher order equations. It also allows for the use of initial conditions to find specific solutions to the equation.

What are the limitations of the reduction of order method?

The reduction of order method can only be used for second order linear homogeneous ODEs. It also does not work for equations with repeated roots or complex roots. In some cases, it may also lead to a more complicated solution compared to other methods.

Can the reduction of order method be used for non-homogeneous equations?

No, the reduction of order method can only be applied to solve homogeneous equations. For non-homogeneous equations, other methods such as variation of parameters or undetermined coefficients must be used.

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