Ensino Superior ⇒ Derivação implícita - aplicações guidorizze Tópico resolvido

[magben]

Seja [tex3]y=e^{-t}.cos(2t)[/tex3]. Verifique que


[tex3]\frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+5y=0[/tex3]

[Cardoso1979]

Observe

Uma verificação:

Temos que

[tex3]y=e^{-t}.cos(2t) \ ( I )[/tex3]

Derivando ( I ) , vem ;

[tex3]y'=[e^{-t}.cos(2t)]'[/tex3]

[tex3]y'=(e^{-t})'.cos(2t)+e^{-t}.[cos(2t)]'[/tex3]

[tex3]y'=(-t)'.e^{-t}.cos(2t)+e^{-t}.(2t)'.[-sen(2t)][/tex3]

[tex3]y'=(-1).e^{-t}.cos(2t)-e^{-t}.(2).[sen(2t)][/tex3]

Logo,

[tex3]y'=-e^{-t}.cos(2t)-2e^{-t}sen(2t) [/tex3]

Ou

[tex3]\frac{dy}{dt}=-e^{-t}.cos(2t)-2e^{-t}sen(2t) \ (II)[/tex3]

Derivando ( I I ) , temos

[tex3]y''=[-e^{-t}.cos(2t)-2e^{-t}sen(2t)]' [/tex3]

[tex3]y''=[-e^{-t}.cos(2t)]'+[-2e^{-t}sen(2t)]' [/tex3]

[tex3]y''=(-e^{-t})'.cos (2t)-e^{-t}.[cos(2t)]'-(2e^{-t})'.sen(2t)-2e^{-t}.[sen (2t)]' [/tex3]

[tex3]y''=(-t)'.(-e^{-t}).cos (2t)-e^{-t}.(2t)'.[-sen(2t)]-(-t)'.2e^{-t}.sen(2t)-2e^{-t}.(2t)'.cos (2t)[/tex3]

[tex3]y''=1.e^{-t}.cos (2t)-e^{-t}.2.[-sen(2t)]+1.2e^{-t}.sen(2t)-2e^{-t}.2.cos (2t)[/tex3]

[tex3]y''=e^{-t}.cos (2t)+2e^{-t}.sen(2t)+2e^{-t}.sen(2t)-4e^{-t}.cos (2t)[/tex3]

Logo,

[tex3]y''=4e^{-t}.sen(2t)- 3e^{-t}.cos (2t)[/tex3]

Ou

[tex3]\frac{d^2y}{dt}=4e^{-t}.sen(2t)- 3e^{-t}.cos (2t) \ (III)[/tex3]


Vamos então substituir ( I ) , ( I I ) e ( I I I ) em [tex3]\frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+5y=0[/tex3], fica;

[tex3]4e^{-t}.sen(2t)- 3e^{-t}.cos (2t)+2.[ -e^{-t}.cos(2t)-2e^{-t}sen(2t)] +5e^{-t}.cos(2t)=[/tex3]

[tex3]\cancel{4e^{-t}.sen(2t)}- 3e^{-t}.cos (2t) -2e^{-t}.cos(2t)-\cancel{4e^{-t}sen(2t)} +5e^{-t}.cos(2t)=[/tex3]

[tex3]- 5e^{-t}.cos (2t) +5e^{-t}.cos(2t)=0[/tex3]. C.q.v.



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