The complete question is
"Find the general solution of the given differential equation
y''-y=0, y1(t)=e^t , y2(t)=cosht
The function [tex]y(t)=e^t[/tex] is the solution of the given differential equation.
The function y(t)=cosht is the solution of given differential equation.
The function is a type of relation, or rule, that maps one input to specific single output.
Given;
[tex]y_1(t) = e^t[/tex]
Given differential equations are,
y''-y = 0
So that,
[tex]y' (t) = e^t, y'' (t) = e^t[/tex]
Substitute values in the given differential equation.
[tex]e^t -e^t=0[/tex]
Therefore, the function [tex]y(t)=e^t[/tex] is the solution of the given differential equation.
Another function;
[tex]y(t)=cosht[/tex]
So that,
[tex]y"(t)=sinht\\\\y"(t)=cosht[/tex]
Hence, function y(t)=cosht is solution of given differential equation.
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