Differential Equations

[Archemeny]

[sanctus]

Looks like homwork, but assuming your equation for y(t) is right (did not check) then it is so simple that I can give you the answer without feeling to compromise your learning curve...
You have found this:

[latex]y(t)=Ae^{-2t}+frac{K}{5}e^{3t}[/latex]

 

Additionally the intial equations tell you that y(0)=0

 

This gives in your found equation:

[latex]0=A{-2\cdot 0}+frac{A}{5}e^{3\cdot 0}[/latex]

 

Which solving for A yields:

[latex] A=\frac{-K}{5}[/latex]

[sanctus]

Sorry for latex tags atm not working, but you should understand what I wrote even if you do not know latex

[Archemeny]

I think the boundary condition shcould be changed to this : system is initially at rest Cause the boundary Condition Y(o) =o tells us only about t=o not (t=o +) , remember We solve this equation For t>o not t=o or to . . .

I must say I found this problem in One of the examples of the book signals and systems by Oppenheim . . .

anyway thanks a lot For your attention Sanctus . . .

[sanctus]

What do you mean by 0+? t tending to zero from the right? If so, why does my approach not work? I mean

[latex]lim_{t\to0+}[/latex] of y(t) is well what I wrote, no?

[Archemeny]

maybe I didn't understand you very well but my main question is: why y(0)= y (0+) ?

[Note that y(0+) is y tending Zero from right ]

[sanctus]

It does not have to be, but if:

[math] lim_{t\to 0+}y(t) =lim_{t\to 0-}y(t)[/math] then you have y(0)=y(0+)=y(0-)

 

And since your y(t) is a continuos function with no singularity the above condition holds.