- #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,

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.

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.