JM00404 Messages 7 Reaction score 0 Thread starter Oct 11, 2005 #1 Please see the PDF attatchment to view the problems. Thank you for your time. Attachments 34.pdf 34.pdf 49.4 KB · Views: 309
arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,165 Reaction score 138 Oct 12, 2005 #2 For 1: What diff.eq does [tex]\psi(x)e^{-a_{1}\frac{x}{n}}[/tex] fulfill?
saltydog Science Advisor Homework Helper Messages 1,590 Reaction score 3 Oct 13, 2005 #3 JM00404 said: Please see the PDF attatchment to view the problems. Thank you for your time. The first one (too late I know but anyway): Let's do a simple one first: [tex]y^{''}+a_1y^{'}+a_2y=0[/tex] or: [tex](D^2+a_1D+a_2)y=0[/tex] or: [tex]f(D)y=0[/tex] Now, use the exponential shift: [tex]e^{cx}f(D)y=f(D-c)[e^{cx}y][/tex] so up there, multiply by: [tex]e^{a_1x/2}[/tex] so: [tex]e^{a_1x/2}f(D)y=f(D-a_1/2)[e^{a_1x/2}\phi]=0[/tex] can you finish it? [tex]f(D-a_1/2)=(D-a_1/2)^2+a_1(D-a_1/2)+a_2[/tex]
JM00404 said: Please see the PDF attatchment to view the problems. Thank you for your time. The first one (too late I know but anyway): Let's do a simple one first: [tex]y^{''}+a_1y^{'}+a_2y=0[/tex] or: [tex](D^2+a_1D+a_2)y=0[/tex] or: [tex]f(D)y=0[/tex] Now, use the exponential shift: [tex]e^{cx}f(D)y=f(D-c)[e^{cx}y][/tex] so up there, multiply by: [tex]e^{a_1x/2}[/tex] so: [tex]e^{a_1x/2}f(D)y=f(D-a_1/2)[e^{a_1x/2}\phi]=0[/tex] can you finish it? [tex]f(D-a_1/2)=(D-a_1/2)^2+a_1(D-a_1/2)+a_2[/tex]