Reduction of order for Second Order Differential Equation

  • #1
ChiralSuperfields
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Homework Statement
Please see below. I am trying to find the reduction of order for a second order ODE using an alternative method than shown in the textbook.
Relevant Equations
Ansatz ##v(t) = e^{rt}##
For this,
1712532117885.png

I tried solving the differential equation using an alternative method. My alternative method starts at

##tv^{''} + v^{'} = 0##
I substitute ##v(t) = e^{rt}## into the equation getting,
##tr^2e^{rt} + re^{rt} = 0##
##e^{rt}[tr^2 + r] = 0##
##e^{rt} = 0## or ##tr^2 + r = 0##
Note that ##e^{rt} ≠ 0##
##tr^2 + r = 0##
##r(tr + 1) = 0##
##r = 0## or ##r = -\frac{1}{t}##

Thus, ##v_1 = e^0 = 1## and ##v_2 = e^{-1} = \frac{1}{e}##

Note that ##v_1 = 1## is a trivial solution since it is just ##x_2 = t = x_1##, however, for ##v_2##, we get ##x_2 = \frac{t}{e}##.

However, ##x_2 = \frac{t}{e}## is just another multiple of ##x_1##. Is it possible to get ##t\log_e(t)## using my method?

Thanks.

Any help greatly appreciated - Chiral.
 
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  • #2
You are assuming v(t)=ert to try to get the solution v(t)=ln(t). Do you see the problem?
 
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  • #3
Your method is invalid. You assume ##e^{rt}## is a solution with constant ##r## and then arrive at ##r = -1/t##. By coincidence ##r=0## makes your ansatz coincide with one of the solutions, but when you find non-constant ##r## you should realize that your ansatz did not work.

Why is this the case? The ansatz ##e^{rt}## is a good ansatz for ODEs with constant coefficients. Your ODE does not have constant coefficients. Instead, it is of Caucy-Euler form.
 
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