Recent content by knockout_artist

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] -...

ok we have to make characteristic equation out of it so I think from this 2d2u/d2x + 1/2Lu = 0 following equation comes 2λ2 + Lλ = 0

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 )...

(sorry missed n in the subject) Hello, How equation 2.2 and 2.3 came about from equation 2.1?

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...

So like this ?