Thread: 2nd Order Diff. Equ. Solution

My goal is to find the general solution of

x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

What we've learned about 2nd order equations is this:

1. How to reduce the order when y or x is absent from the equation
2. How to solve equations in the form y'' + py' + qy = 0, where p and q are constants
3. How to solve equations in the form y'' + py' +qy = ke^(ax), k*sin(b*x) + r*cos(b*x), or a + b*x + c*x^2, where a, b, c, k, and r are constants.
4. How to find a second solution to the equation y'' + p(x)*y' + q(x)*y = 0, where p(x) and q(x) are functions of x, when one solution is already known.
5. How to find a particular solution to the equation y'' + p(x)*y' + q(x)*y = r(x), when the general solution to y'' + p(x)*y' + q(x)*y = 0 is known.

The problem is, I don't know how to reduce

x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

to an equation in any of the forms above. That is, I don't even know how to find the homogeneous equation, because I can't separate x out of the right side entirely.

At this level of understanding, should I be using intuition to find one solution, or is there a way to use substitution, or what?

It looked like I could factor the left side, up until the 2*y part.

Anyway, I'm stumped. Any help is much appreciated.

Originally Posted by primasapere

My goal is to find the general solution of

x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

What we've learned about 2nd order equations is this:

1. How to reduce the order when y or x is absent from the equation
2. How to solve equations in the form y'' + py' + qy = 0, where p and q are constants
3. How to solve equations in the form y'' + py' +qy = ke^(ax), k*sin(b*x) + r*cos(b*x), or a + b*x + c*x^2, where a, b, c, k, and r are constants.
4. How to find a second solution to the equation y'' + p(x)*y' + q(x)*y = 0, where p(x) and q(x) are functions of x, when one solution is already known.
5. How to find a particular solution to the equation y'' + p(x)*y' + q(x)*y = r(x), when the general solution to y'' + p(x)*y' + q(x)*y = 0 is known.

The problem is, I don't know how to reduce

x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

to an equation in any of the forms above. That is, I don't even know how to find the homogeneous equation, because I can't separate x out of the right side entirely.

At this level of understanding, should I be using intuition to find one solution, or is there a way to use substitution, or what?

It looked like I could factor the left side, up until the 2*y part.

Anyway, I'm stumped. Any help is much appreciated.

This is Cauchy's ODE, and you make the change of variable:

t=ln(x),

or equivalently:

x=exp(t).

Which transforms the quation in to a 2nd order ODE with constant
coefficients. (see here)

RonL

Originally Posted by primasapere

My goal is to find the general solution of

x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

Actually it is simpler than CaptainBlank said.
It is the Euler equation.

Solve, the homogenous equation,
k(k-1)-2k+2=0
Thus,
k^2-3k+2=0
k=[3+/-sqrt(5)]/2

Thus, the homogenous solution is,
y=c_1 x^(3/2) cos(\sqrt(5)/2 ln |x|)+c_2 x^(3/2)sin(\sqrt(5)/2 \ln|x|)

Now for the particular solution you can pretend it is of the form
y=Axe^{-x}

And take it from there.

If not use Largrange's varaition of parameters techiqnue