What is the issue with the last term in the conversion from v(t) to y(t)?

[Shackleford]

For #3, when going from the v(t) to y(t), I wasn't sure what to do with the
C. When you get to the y(t), the last term is C / t^2. When you put in the
initial condition y(0), you get an indeterminate expression C / 0.

http://i111.photobucket.com/albums/n149/camarolt4z28/3.jpg

 

[foxjwill]

For #3, when going from the v(t) to y(t), I wasn't sure what to do with the
C. When you get to the y(t), the last term is C / t^2. When you put in the
initial condition y(0), you get an indeterminate expression C / 0.

http://i111.photobucket.com/albums/n149/camarolt4z28/3.jpg

There isn't much you can do. However, C/0 is not indeterminate! It's simply undefined. "Indeterminate" only has meaning when the expression is inside the argument of a limit.

 

[HallsofIvy]

You have y'= a(t)y+ f(t) and assert that the integrating factor is
[tex]e^{\int a(t)dt}[/itex]<br /> That is incorrect. The formula is for a d.e. of the form y'+ a(t)y= f(t) so you have the sign wrong. The equation y'= -(2/t)y+ t-1 is equivalent to y'+ (2/t)y= t- 1. The integrating factor is <br /> $$e^{\int 2/t dt}= e^{2 ln|t|}= t^2$$.<br /> <br /> Of course, you are still going to have a problem at t= 0 because one of the coefficients of your d.e. is not defined at t= 0.[/tex]

 

[Shackleford]

You have y'= a(t)y+ f(t) and assert that the integrating factor is
[tex]e^{\int a(t)dt}[/itex]<br /> That is incorrect. The formula is for a d.e. of the form y'+ a(t)y= f(t) so you have the sign wrong. The equation y'= -(2/t)y+ t-1 is equivalent to y'+ (2/t)y= t- 1. The integrating factor is <br /> $$e^{\int 2/t dt}= e^{2 ln|t|}= t^2$$.<br /> <br /> Of course, you are still going to have a problem at t= 0 because one of the coefficients of your d.e. is not defined at t= 0.[/tex]

$$<br /> Hm. I thought it didn't have to be in standard form.<br /> <br /> How do you get around the undetermined expression?$$