Can a Substitution Solve a Linear 2nd Order ODE with Non-Constant Coefficients?

kyin01
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Homework Statement


y''x^{2} + 4xy' -y = ln(x)

The Attempt at a Solution


-I considered the quadratic characteristic equation, but it won't work because of the x^2
-I also tried variation of parameters.
so i have v = y' and v'=y'' but i have no idea what i would sub when I get to y.

any ideas?
 
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Hi kyin01! :smile:

Try changing x

look for a u(x) such that the equation becomes something simple in dy/du and d2y/du2. :wink:
 
kyin01 said:

Homework Statement


y''x^{2} + 4xy' -y = ln(x)

The Attempt at a Solution


-I considered the quadratic characteristic equation, but it won't work because of the x^2
-I also tried variation of parameters.
so i have v = y' and v'=y'' but i have no idea what i would sub when I get to y.

any ideas?

I think the standard way is to first find the solutions of the assoc. homogeneous DE y''x^{2} + 4xy' -y =0 using Cauchy Euler DE by letting y=x^{m}.

Then use variation of parameter etc. to find the particular solution.
 
Yes, although that can make the calculations pretty complicated. As tiny-time suggested, the substitution u= ln(x) will change any "Euler-type" or "Equipotential" equation to a very simple equation with constant coefficients.

I rather suspect that if kyin01 were to check his recent notes or textbook where this problem is given, he would find a discussion of "Euler-type" or "Equipotential" equations. (The name varies from textbook to textbook.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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