Homework Statement
[/B]
For differential equation of the form
## y''- y = 0 ##
BC is
## y(1) = B ##
which usually have general solution
## y(x) = C1 e^x + C2 e^{-x} ##
But this manual I am reading always want to go with general solution
## y = C1 \cosh(x) + C2 \sinh( x) ##
I assume...
Homework Statement
Taking derivative of 3 term product.
## \frac{d}{dx} (3x^3 y^2 y'^2) ##
Homework Equations
I read that
(abc)' = (ab)c' + (bc)a' + (ca)b'
The Attempt at a Solution
## 9x^2(y^2y'^2) + 6x^3yy'^2 + 6x^3y^2y' ##
is this correct ?
I am only member of to math websites.
Other one is not even close to what I am talking about.
Any ways I will try to find one of my old thread and post it here.
thank you.
sigma button brings back symbols only.
the editor I am speaking of and i have used in past few months would pop open a small window.
In that window it will say "do not use $ $ "
and one editing is done and I press " insert" it will put equation in latex form and it will wrap it in
<tex>...
Perhaps its called some thing else.
But I have used it with in last few month.
Please check this link for few months ago which has some editor work in it.
https://www.physicsforums.com/threads/square-root-differential-problem.920046/#post-5802507
Good Morning,
I don't see latex editor or latex equation inserting tool on the editing page any more, that's when I am creating a thread.
Has any this changed ?
Thank you
Homework Statement
Homework Equations
How do we get y(x) ?
should y'(x) in the sqrt not cancel the y/(x) on the RHS ?
Does y(x) comes back because of integrating 1 with dy ?
The Attempt at a Solution
w(x)y'(x) = A (1 +y'(x)2)1/2
(w(x)y'(x) )2 = ( A (1 +y'(x)2)1/2 )2
w(x)2y'(x)2 = A2...
Homework Statement
I am little confused. Value of used to be
sin(0) = 0 = cos(1)
sin(1) = 1 = cos(0)
Homework EquationsThe Attempt at a Solution
Now when I use google calculator I use rad or degree I do not get solid 1 and 0.
for example
sin(0)= 0 deg =0rad
sin(1)= .841 deg = 0.017 rad...
Homework Statement
x2 d2y/dx2 + 3x dy/dx + 5y = g(x)Homework Equations
How do we find Characteristic equation for it.
The Attempt at a Solution
x2λ2 + 3xλ + 5 = 0
λ1 = 1/2 [-x2 + √ (x4 + 20 ) ]
λ2 = 1/2[ -x2 - √(x4 + 20) ]
I used 1/3 -/+ a √(a2 + 4b)
where
a = x2
b = 5
Homework Statement
d2u/d2x + 1/2Lu = 0 where L is function of x
Homework Equations
I am try to find solutions y1 and y2 of this equation.
The Attempt at a Solution
y = [cos √(L/2) x] + [sin √(L/2) x]
y' = - [√(L/2) sin √(L/2) x] + [ √(L/2) cos √(L/2) x]
y'' = -[(L/2) cos √(L/2) x] -...
Hi,
This is equation I need to find solutions for
d2u/d2x + 1/2Lu = 0 where L(x)
I understand we can remove fraction from second term.
2 [d2 u/d2x ] + Lu = 0
now how do I find solution of this equation ?
How do we deal with L ? because usually we have Y'(dy/dx or in this case du/dx )...
So in the book it says expend function ƒ in ε to get following.
ƒ=√ (1 + (α + βε)2) = √ (1 + α2) + (αβε)/√ (1 + α2) + (β2ε2)/2 (1 + α2)3/2 + O(e3)
When I expend I get(keeping ε = 0)
ƒ(0) = √ (1 + α2) -->first term
ƒ'(0) = (αβ)/√ (1 + α2) --> sec term with gets multiplied by ε
for third...