English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Starting with d^2u/dx=du/dy or written u(xx)=u(y)
y>0, 0 u(x,0)=1, u(o,y)=u(a,y)=0
Taking the Laplace transform of u with respect to y gives:
L[u(xx)]=su-1, with the left hand side being just the ordinary differential equation as x is not affected by the Laplace.
The general solution is:
u(x,s) = c(1)e^(xSQRT(s)) + c(2)e^(-xSQRT(s)) + c(3),
which could also be written in terms of cosh and sinh.
How do I find c(3) using the method of undetermined coefficients?
Please try to explain it out. Many of my questions are answered by pulsing brainiacs from another dimension who tend to make succinct responses that they understand but please I am not so smart and have to trudge through step by step. Thanks.

2007-12-07 01:57:13 · 2 answers · asked by entropic v 3 in Science & Mathematics Mathematics

2 answers

Let me write U for L[u]

U'' - sU = -1

When the right side is a polynomial, try a general polynomial of the same order as a particular solution. Here you would just try U = A, A a constant. You get U = 1/s. Then the general solution for U is 1/s + the general solution for the homogeneous problem, which you have found.

2007-12-07 03:18:40 · answer #1 · answered by Ron W 7 · 0 0

y''-4y=0 y= c_1 *e^2x +c_2 e^-2x finding one answer of the non homogeneous equation y= =e^x(acos x+b sinx) y'= e^x(-asinx+b cosx)+e^x(acosx+bsinx) = e^x[(b-a)sinx +(a+b)cosx] y'' = e^x[(b-a)cos x-(a+b)sinx]+e^x[(b-a)sin x+(a+b)cos x]0 e^x[2bcosx-2asin x] 2b-4a=a million -2a-4b=0 a=-2b 10b= a million b=a million/10 a=-a million/5 y= e^x(-a million/5 cosx +a million/10 sinx) +C_1e^2x+C_2e^-2x y(0)=-a million/5 +C_1 +C_2 =0 y'(0) = 2 C-a million -2C-2 + a million/10 -a million/5 =2 you could sparkling up the gadget for C-a million and C_2

2016-12-10 15:29:21 · answer #2 · answered by ? 4 · 0 0

fedest.com, questions and answers