Homework Statement
Using Laplace transform to solve 4th order DE with a delta dirac forcing function. Has a tricky denominator, I just need a clue.
y^{(4)} - y = \delta (t-2)
__
IV's y'''(0)=0 , y''(0)=0 , y'(0)=0 , y(0) = 0
Homework Equations
I am asked to convert the solution...
Yeah, it's 9. Thanks for pointing out my mistake there, but also that's not the problem that I am having. THe thing is that the 9 has a factor of Y(s) so I have no choice but to return to the original equation as far as I can see. That's what is bothering me what can I do about it?
Y(s)s^2 +...
Ok I feel stupid, I was thinking that completing the square would give me Y(s) terms on the RHS which I would have to divide out but I just tried it and now I don't know why I was thinking that.
I have factors but now I have a denominator with a dangling 6 that I don't know what to do with...
Homework Statement
A 2nd order DE with a Delta Dirac forcing function, I have been asked to solve in my DE course.
y'' + 2y' + 10y = \delta(t-1)
Also given are IV's y'(0)=0, y(0)=0
Homework Equations
At this stage in my course I am being asked to only use the method of partial...
It's good to know I'm not alone. :-p
How long I've been working with quadratics is; I missed out on the High school preparation and have been factorising them in college level problems for two years of part time study. I said above that I am *supposed* to be doing it like my signature, but I...
Thankyou Bob. That song is AWful yes but it could be helpful.
On what you wrote above, also thankyou, perhaps my elementary arithmetic skills leave something to be desired. And that is my main trouble with it. I have tried practising superfluously as per the adage that it's the only way...
Why? How bad is my terminology ?
Yeah I am. Do you have any handy factorising tips.
Heres what I can do:
if I have 2x^2 - 3x - 2 then I can see that my factors need to be (x+a)(x-b) to get -2, and I know that a and b must have a product of -2 and 2x*b + x*a will need to have a sum of...
Hi.
It's late in the game for me to still be banging my head over this one, so I was hoping for some priceless tips from those who are doing it in their sleep.. PLeeeasse :redface:
I know I should remember the Quadratic formula better than I do, but I was encouraged not to resort to it...
Thanks Galileo for your reply, I worked out my mistake before I got here it was just a stupid mistake of adding and adding dumb really. :rolleyes:
I just wanted to say that your explanation has helped me the most to grasp the concepts so thanks Reaallly much. o:) :smile:
I can see that:
5(5^k-4k+15)+ 16k-64
Is equal to P(k) + 16(n element of reals)
But I m not sure I understand how it was induced into the equation.
Here is where I am stuck.
I've found:
5 \cdot 5^k-4k +11
naturally.
The difference between 5\cdot 5^k and (5^k-4k+15)...
Thanks for that Saltydog. I could see easily that that was true, but I was not sure what rigorous proof was required to satisfy IVT.
Thankyou again, I see that is one way that I can say it exists before I find it, but I am not sure I will naturally draw this conclusion in a test...
Could someone please explain to me how I can use the intermediate value theorem to convince myself that m where
m = x^5 - 1 = x^3
exists. I have deduced that I can work in interval [1,2] using Newtons method where
x^5 - 1 - x^3 = 0
to find the answer. I even have the answer...
Just the validity of transferring y to differentiate both sides, I wasn't sure that I was differentiating the same equation, probably over thinking it.
Thanks heaps though because I think I am starting to become more capable with it but I don't want to be overconfident in mistakes.
:cool:
To find dy/dx for
y = ln(sin^{-1}(x))
I did this:
y = ln(sin^{-1}(x))
so
e^y = (sin^{-1}(x))
and
e^y \frac {dy}{dx} = \frac {1} {\sqrt{1-x^2}}
then
\frac {dy}{dx} = \frac {1}{\sqrt {1-x^2} * e^y}
= \frac {1}{\sqrt {1-x^2} *...