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2nd order ODE

It's been a while since I've done one of these which could be why I'm struggling with this question.

Find the general solution of:

\displaystyle \frac{d^2 y}{d t^2}+6\frac{d y}{d t} +9y=4e^{-3t}

I found the complementary function:

\displaystyle y=(A+bt)e^{-3t}

I've tried a few different trial solutions for the particular integral but none of them seem to work. Can someone help me out?

Reply 1

deadgenius

Did you remember that when the CF is similar to the right-hand side of the equation you have to differentiate ke^(-3t) rather than just e^(-3t).

Reply 2

0-)

deadgenius

Did you remember that when the CF is similar to the right-hand side of the equation you have to differentiate ke^(-3t) rather than just e^(-3t).

Do you mean kte^(-3t)? ke^(-3t) would be the regular trial solution.

I've tried ke^(-3t), kte^(-3t) and kt^2 e^(-3t).

Reply 3

deadgenius

Haha yeah. I knew there was a trick with k in there somewhere. I'm a bit rusty myself. Ok, I'll give this question a proper go. Hang on and I'll come back in about 10 minutes.

Reply 4

motoroller

0-)

I've tried...

Surely those should do the trick? Those are the functions you should use to get the particular integral...

Reply 5

vaseand

If the particular solution you are trying to use is similar to the homogenous solution, try using the guess - y= kte^-3t instead of ke^-3t. Should work if you do that

Reply 6

0-)

vaseand: I've already tried that.

I tried solving it in MAPLE and the PI involved division by zero. Maybe there is no particular integral...?

Reply 7

deadgenius

Ok I did what you did and ended up with nothing. But I tried using

y = kt^{3}e^{-3t}

and that worked. I got

k = \frac{2}{3t}

. I don't know if that's right, but it's the only answer I have!

Reply 8

quakeeem

try

Unparseable latex formula:

y=kt^2 e^-^3^t

again

I have a feeling that you may've cancelled something by accident.

nb (because it does work)

Reply 9

Swayum

deadgenius

Ok I did what you did and ended up with nothing. But I tried using

y = kt^{3}e^{-3t}

and that worked. I got

k = \frac{2}{3t}

. I don't know if that's right, but it's the only answer I have!

But then surely it's just

kt^{3}e^{-3t} = \frac{2}{3t}t^{3}e^{-3t} = \frac{2t^2e^{-3t}}{3}

?

These types of 2nd order ODEs should be soluble by using at most t^2 * e^whatever. I haven't tried it, but I've seen a proof that it should work.

Reply 10

deadgenius

Ah yes. quakeem is right. I must have done the same thing as you - cancelled one too many things!

Reply 11

quakeeem

deadgenius

Ok I did what you did and ended up with nothing. But I tried using

y = kt^{3}e^{-3t}

and that worked. I got

k = \frac{2}{3t}

. I don't know if that's right, but it's the only answer I have!

k is a constant, it shouldn't depend on

t

.

I'll spoiler my working:

Spoiler

Reply 12

deadgenius

Swayum

But then surely it's just

kt^{3}e^{-3t} = \frac{2}{3t}t^{3}e^{-3t} = \frac{2t^2e^{-3t}}{3}

?

These types of 2nd order ODEs should be soluble by using at most t^2 * e^whatever. I haven't tried it, but I've seen a proof that it should work.

Yeah I thought that was a bit odd, and looking back at my working out for t^2 * e^(-3t) I realised I made a mistake. Whoops!

Reply 13

deadgenius

quakeeem

k is a constant, it shouldn't depend on

t

.

I'll spoiler my working:

Spoiler

Yeah I did get that. I just did it far too quickly and just cancelled out everything lol.

Reply 14

0-)

quakeeem

k is a constant, it shouldn't depend on

t

.

I'll spoiler my working:

Spoiler

Oops - I completely ignored the first term in y'' for some reason. I have the answer now.

Going off the point: Does anyone know how to set up MAPLE so that it knows that ln(e) is 1?

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