Recent content by dp182

Homework Statement given that kλ2-ρcpuλ-ρcpωi=0 plug into the quadratic formula and get out an equation that looks like this λ=α+iβ±γ√(1+iδ) where α,β,γ,and δ are in terms of ρ,cp,u,k, and ω Homework Equations (-b±√b2-4ac)/2a kλ2-ρcpuλ-ρcpωi=0 λ=α+iβ±γ√(1+iδ) The Attempt at a...

I believe T is constant. so I tried using variation of parameters to solve (Uinf=X) X''-4X=-4T and came up with a particular solution of c1e2x+c2e-2x-T then solved for the constants using boundary conditions. is this method right as all the example I have seen involve just integrating it twice.

Homework Statement determine the steady state equation for the given heat equation and boundary conditions Homework Equations Ut=Uxx-4(U-T) U(0,T)=T U(4,T)=0 U(x,0)=f(x) The Attempt at a Solution I put Ut=0 so 0=UInf''-4(Uinf-T) then once I tried to integrate I ended up with a...

ya but I brought out a 4/n2pi2 so its not -1 its -4/n2pi2

but wouldn't F(0) be 4cos(0)/n2pi2 which is just 4/n2pi2

Homework Statement let f(x)={0;-2\leqx\leq0. x;0\leqx\leq2 find a0 an bn given the period is 4 Homework Equations a0=1/L\intf(x)dx an=1/L\intf(x)cos(n\pix/L) bn=1/L\intf(x)sin(n\pix/L) The Attempt at a Solution so I can get a0 = 1 but I run into trouble with an. so I plug...

Homework Statement 2xy''-x(x-1)y'-y=0 about x=0 what are the roots of the indicial equation and for the roots find the recurrence relation that defines the the coef an Homework Equations 2xy''-x(x-1)y'-y=0 about x=0 assuming the solution has the form y=\Sigmaanxn+r...

So If I solve for an+2 I get an+2=n(n+1)an+1+an(2-n)/(n+2)(n+1) so to get my terms up to x6 I will just input values of n, so my series will be in terms of a0and a1?

Homework Statement consider the initial value problem (1-x)y,,+xy,-2y=0 find the series solution up to the term with x6 Homework Equations (1-x)y,,+xy,-2y=0 The Attempt at a Solution assuming the answer has the form \Sigmaanxn that gives y,,=\Sigmananxn-1 and...

Nvm I am an idiot lol thanks a lot you've really helped

so then the equation for am+2 would be am+2=(-m(m-1)+m-4)am/(m+2)(m+1) which would be the same one I got before

when you took the answer to the ode y=\sumanxn and then derived it to get y'=\sumanxn-1 shouldn't you bring down the n as a constant? and if not then wouldn't your solution for am+2=am?

That is the same as what I did except there are no constants when you took the derivative's of the series how come you don't add any new terms when you took your derivatives

from the beginning: since the answer is of the form y=\Sigmaanxn so the first and second order derivatives are y'=\Sigmananxn-1 and y''=\Sigman(n-1)anxn-2 so plugging those into the original equation we get (2+x2)\Sigman(n-1)anxn-2-x\Sigmananxn-1+4\Sigmaanxn and from that equation you can see...

I find two series with a(0) and a(1) but its wrong and no matter how many times I try to equate my equation for a(n+2) i get the same equation I posted so I'm not sure where I am screwing up