solving by substituting a complex function

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• Dec 2nd 2010, 06:47 AM
transgalactic
solving by substituting a complex function
i need to solve
$x^2y''-xy'+y=4x^2$

by saying that $x=e^t$
and Y(t)=y(x(t))

then i need to find from them expresions for \fracxdy/dx and $x^2d^2y/dx^2$ which are expresions of dY/dt similar type

i tried like this:
$x=e^t$
$\frac{dx}{dt}=e^t$
Y(t)=y(x(t))
$\frac{dY(t)}{dt}=\frac{dy(x)}{dx}\frac{dx(t)}{dt}$
$\frac{dY(t)}{dt}=\frac{dy(x)}{dx}x$
$\frac{dY^2(t)}{dt}=\frac{d^2y(x)}{dx^2}x^2+\frac{d y(x)}{dx}$

$y'x=\frac{dY(t)}{dt}$
$y''x^2=\frac{dY^2(t)}{dt}-y'$

so
$\frac{dY^2(t)}{dt}-y'-\frac{dY(t)}{dt}+y=4x^2$

what now
?
• Dec 2nd 2010, 07:14 AM
Ackbeet
I'm not sure you're doing the second derivative properly. I agree with the first derivative, but you have to use the chain rule on the second derivative as well. See here.
• Dec 2nd 2010, 07:21 AM
transgalactic
i replaces e^t with x
so when i do the chain rules in the second derivative dx(t)/dt=e^t=x
that how i got x^2
i cant see the mistake
in general i dont see how this method works
• Dec 2nd 2010, 07:30 AM
Ackbeet
You've correctly got

$\dot{Y}=\dfrac{dy}{dx}\,\dfrac{dx}{dt}=e^{t}y',$ and hence $y'=e^{-t}\dot{Y}.$

Next, you compute

$\ddot{Y}=\dfrac{d}{dt}\left(e^{t}\,\dfrac{dy}{dx}\ right)
=e^{t}\dfrac{dy}{dx}+e^{t}\dfrac{d}{dt}\,\dfrac{dy }{dx}
=\dot{Y}+e^{t}\left(\dfrac{d}{dx}\dfrac{dy}{dx}\ri ght)\dfrac{dx}{dt}
=\dot{Y}+e^{2t}y''.$

Thus,

$\ddot{Y}-\dot{Y}=e^{2t}y'',$ and therefore

$y''=e^{-2t}(\ddot{Y}-\dot{Y}).$

$e^{2t}(e^{-2t}(\ddot{Y}-\dot{Y}))-e^{t}(e^{-t}\dot{Y})+Y=4e^{2t}.$

When you simplify, lots of nice things happen. Does this make sense?
• Dec 2nd 2010, 07:34 AM
transgalactic
my prof said that x=e^t
and when i see dx/dt i replace it with x
because dx/dt=e^t=x

so there isnt supposed to be e^t in your solution
• Dec 2nd 2010, 07:40 AM
Ackbeet
Well, continue simplifying, and substitute back in so you just have x's. It's six of one, a half-dozen of the other.

Incidentally, this method of solution is inferior to the usual Cauchy-Euler method, because with this solution, you are artificially constraining x to be positive, whereas that is not the case with the usual Cauchy-Euler method.
• Dec 2nd 2010, 07:46 AM
transgalactic
$(\ddot{Y}-x^2\dot{Y}))-(\dot{Y})+xY=4x^2$
$\ddot{Y}+(-x^2-1)\dot{Y}+xY=4x^2$

ok i did the substitution
how it became easier
?
• Dec 2nd 2010, 07:48 AM
Ackbeet
That is incorrect. How did you get that from my post # 4?
• Dec 2nd 2010, 07:55 AM
transgalactic
$
e^{2t}(e^{-2t}(\ddot{Y}-\dot{Y}))-e^{t}(e^{-t}\dot{Y})+Y=4e^{2t}$
$
e^t=x
$

$
e^2t=x^2

$

4 members canceled each other
$(\ddot{Y}-x^2\dot{Y}))-(\dot{Y})+xY=4x^2$
$\ddot{Y}+(-x^2-1)\dot{Y}+xY=4x^2$
i cant see my mistake

and its still a second order equation with x's as coefficient
i cant see how it got easier
• Dec 2nd 2010, 07:59 AM
Ackbeet
You made two mistakes. The first one was in not noticing that the $e^{-2t}$ multiplies both terms in the expression $\ddot{Y}-\dot{Y}.$ The second mistake is the mysterious $x$ that multiplies the $Y$ as your last term on the LHS. That seems to come out of thin air. The $e^{t}$ multiplying the $e^{-t}\dot{Y}$ term does NOT multiply the $Y$ term.

Perhaps both of these mistakes are a result of not noticing the parentheses carefully enough? You absolutely have to pay attention to details like parentheses if you're going to avoid mistakes.
• Dec 2nd 2010, 08:19 AM
transgalactic
ok here i go carefull on every step
first i open the parentheses and put x^2 instead of $e^{2t}$
$\ddot{Y}-2\dot{Y}+Y=4x^2$

but Y is a function of t
not x
how to go further
• Dec 2nd 2010, 08:22 AM
Ackbeet
Correct. Now, here's where I would differ with your professor, perhaps. You should have the exponentials on the RHS, not x^2. Then you've got a second-order linear non-homogeneous ode with constant coefficients. Turn the crank.
• Dec 2nd 2010, 08:33 AM
transgalactic
ok but how this method differs the y=x^a substitution method
?

my prof said that the exponent method is better because it solves equations that
x^a cant

in wikipedea they dont say anything about the difference between them

on what type i need to use what method
?
• Dec 2nd 2010, 08:40 AM
Ackbeet
I'm not overly familiar with the differences between the two methods, other than the observation I made before. If you have a bona-fide Cauchy-Euler equation, then the x^a method should solve it. The exponential substitution might work for other types of DE's, but as I said, you're restricting yourself to positive x's. That might not matter so much if your x is really a t, and you're only interested in positive time.
• Dec 8th 2010, 08:58 AM
transgalactic
cant understand the transition from
$e^{t}\dfrac{d}{dt}\,\dfrac{dy}{dx}$
to
$e^{t}\left(\dfrac{d}{dx}\dfrac{dy}{dx}\right)\dfra c{dx}{dt}
$
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