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Given that y(t)=c1e4t+c2e−4ty(t)=c1e4t+c2e−4t is a solution to the differential equation y′′−16y=0y′′−16y=0, where c1c1 and c2c2 are arbitrary constants, find a function y(t)y(t) that satisfies the conditions

Respuesta :

Solution :- Given differential equation

[tex]y′′−16y=0[/tex]

 So the characteristic equation will be

[tex]r^{2} -16=0


⇒r^{2} =16


⇒[tex]r_{1}=4 \\or \\ r_{2}=-4[/tex]

so, the  required  function will be

[tex]y=c_{1}e^{r_{1}t}+c_{1}e^{r_{2}t}[/tex]

⇒ [tex]y=c_{1}e^{4t}+c_{1}e^{-4t}[/tex] ia solution of the  given differential equation  

where [tex]c_{1} \\and \\c_{2}[/tex] are constants.

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