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Mathematics 12 Online
OpenStudy (anonymous):

GIVE THE COMPLETE SOLUTION,FOR THE SECOND ORDER IVP : S(n)=2S(n-1)-S(n-2) S(0)=1,S(1)=4, (WHERE n>=2)

OpenStudy (anonymous):

write out a few terms... you'll see you get 3k+1

OpenStudy (blockcolder):

Do you need a formal solution?

OpenStudy (anonymous):

yes pls

OpenStudy (blockcolder):

\[S_n=2S_{n-1}-S_{n-2}\] The characteristic equation and roots are: \[r^2=2r-1\\ r^2-2r+1=0\\ (r-1)^2=0\\ r=1\] Thus, the solution is of the form: \[S_n=\alpha_1 1^n+n\alpha_21^n=\alpha_1+n\alpha_2\] Plugging in the initial values, we get: \[\begin{cases} S_0=1=\alpha_1\\ S_1=4=\alpha_1+\alpha_2 \end{cases}\\ \alpha_2=3\] So the solution is S(n)=1+3n.

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