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Differential Equations


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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]

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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 . . .

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