Recent content by roughwinds

Homework Statement I'm unsure of what exactly is changing the heat transfer direction in the triangular fin. Homework Equations $$q_{x} = -kA(x)\frac{dT(x)}{dx} (1)$$ $$q_{x+dx} = -kA(x)\frac{dT(x)}{dx} - k\frac{d}{dx}[A(x)\frac{dT(x)}{dx}] (2)$$ $$dq_{conv} = h(x)dS(x)P[T(x) - T_{∞}] (3)$$...

$$a_0 r^2 + [\alpha a_0 +(r+1)^2 a_1]x = 0$$ $$a_0r^2 = 0$$ $$r = 0$$ $$\alpha a_0 +(r+1)^2 a_1 = 0$$ $$a_1 = \frac{-\alpha a_0}{(r+1)^2} = -\alpha a_0$$ $$a_n = \frac {-(αa_{n-1}+βa_{n-2})}{(n+r)^{2}}$$ $$a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n^{2}}$$ $$a_2 = \frac {(α^{2}+β)a_0}{4}$$ $$a_3 =...

Frobenius attempt: y = \sum_{n=0}^\infty a_n x^{n+r} y' = \sum_{n=0}^\infty a_n(n+r) x^{n+r-1} y'' = \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r-2} x²y'' = \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r} = a_0r(r-1)x^{r} + a_1r(r+1)x^{r+1} + \sum_{n=2}^\infty a_n(n+r)(n+r-1) x^{n+r} xy' =...

Homework Statement Solve for xy'' + y' +αy + βxy = 0 α and β are constants The Attempt at a Solution What I initially had in mind was: xy'' + y' +αy + βxy = x²y'' + xy' +αxy + βx²y = 0 y = \sum_{n=0}^\infty a_n x^{n} xy = \sum_{n=0}^\infty a_n x^{n+1} = \sum_{n=1}^\infty a_{n-1} x^{n} = a_0x...

-\frac{d(s²L(y))}{ds} = -2sL(y) - \frac{s²d(L(y))}{ds} -2sL(y) - \frac{s²d(L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0 - \frac{s²d(L(y))}{ds} - sL(y) -\frac{4dL(y)}{ds} =0 \frac{s²d(L(y))}{ds} + sL(y) +\frac{4dL(y)}{ds} =0 \frac{d(L(y))(s²+4)}{ds} + sL(y) =0 Thanks, I initially solved as if s²...

It's supposed to be -\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0 fixed it on the original post.

Homework Statement Use Laplace transform to solve the following ODE Homework Equations xy'' + y' + 4xy = 0, y(0) = 3, y'(0) = 0 The Attempt at a Solution L(xy'') = -\frac{dL(y'')}{ds} L(4xy) = -\frac{4dL(y)}{ds} L(y'') = s²L(y) - sy(0) - y'(0) = s²L(y) -3s L(y') = sL(y) - sy(0) - y(0) =...

Wouldn't it have to be x²θ′′+xθ′−mθ=0 for it to be an Euler differential equation? I tried to do power series but I got stuck in the end: y=\sum_{n=0}^\infty a_k x^k θ'=\sum_{n=1}^\infty ka_k x^{k-1} θ''=\sum_{n=2}^\infty k(k-1)a_k x^{k-2} \sum_{n=2}^\infty k(k-1)a_k x^{k-1} +...

I don't think I expressed myself properly before but anyway, finally managed to get it in the way I wanted: http://i.share.pho.to/0a247f78_o.jpeg http://i.share.pho.to/57739467_o.jpeg T(x) - T_e = θ(x) m = \frac{h_e(L/e)(1+(e/L)²)^{1/2}}{k} xθ'' + θ' - mθ = 0 Just need to solve that now.

So this is correct ? q_x = -kA(x)\frac{dT(x) }{dx} q_{x+dx} = -kA(x)\frac{dT(x) }{dx} - k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²} dq_{conv} = h_eP(T(x)-T_e)dx -kA(x)\frac{dT(x) }{dx} = -kA(x)\frac{dT(x) }{dx} - k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²} + h_eP(T(x)-T_e)dx...

I'm trying to deduce the differential equation for temperature for a triangular fin: I know that for a rectangular fin, such as: I can do: Energy entering the left: q_x= -kA\frac{dT(x)}{dx} Energy leaving the right: q_{x+dx} = -kA\frac{dT(x)}{dx} - kA\frac{d² T(x)}{dx²}dx Energy lost by...