Hi there, I was just trying to perform a change of variable on a differential equation as shown below. The original second order ODE is written in terms of a dependent variable ## \theta ## and independent variable ##q##. I have used the expression ## q = \sqrt{\frac{z\tau'}{LT_R}} ## and ##\tau' = T'(1-e^{\frac{-\tau}{T'}}) ## (where ##z## and ##\tau## are also independent variables, while ##L, T',## and ##T_R## are known constants). I am only considering the case of ##z = 0.01L## and am thus only solving this equation for a varying ##\tau##. (In this case, ##\tau## is a purely real and positive number.) I have posted my work below to transform the second order ODE into a function of ##t## instead of ##q##. (I may have used ##t## and ##\tau## interchangeably out of bad habit--my apologies if this causes confusion.) If you could possibly check it to see if there are any errors, that would be greatly appreciated as I'm still a little uncertain if I've applied the chain rule correctly in certain parts.(adsbygoogle = window.adsbygoogle || []).push({});

(My apologies if this is the wrong section to post such question, although this is not homework.)

The final solution I get (it is partially cut in the last image) is:

$$ 100e^{\frac{2t}{T''}}(1-e^{\frac{-t}{T''}})[(1 + \frac {e^{\frac{-t}{T''}}}{2(1-e^{\frac{-t}{T''}})}) \frac {d\theta}{dt} + T'' \frac {d^2\theta}{dt^2} ] + 50e^{\frac{t}{T''}}\frac {d\theta}{dt} = sin(\theta) $$

which can also be expressed as:

$$ 100T''(e^{\frac{2t}{T''}}-e^{\frac{t}{T''}}) \frac {d^2\theta}{dt^2} + 50e^{\frac{t}{T''}}[(e^{\frac{t}{T''}}-1)(2+ \frac {1}{e^{\frac{t}{T''}}-1}) + 1]\frac {d\theta}{dt} = sin(\theta) $$

which is equivalent to:

$$ 100T''(e^{\frac{2t}{T''}}-e^{\frac{t}{T''}}) \frac {d^2\theta}{dt^2} + 100e^{\frac{2t}{T''}}\frac {d\theta}{dt} = sin(\theta) $$

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# I Change of variable (2nd order ODE)

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