Respuesta :
Answer:
For First Solution: [tex]y_1(t)=e^t[/tex]
[tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.
For 2nd Solution:[tex]y_2(t)=cosht[/tex]
[tex]y_2(t)=cosht[/tex] is the solution of equation y''-y=0.
Step-by-step explanation:
For First Solution: [tex]y_1(t)=e^t[/tex]
In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.
[tex]y_1(t)=e^t[/tex]
First order derivative:
[tex]y'_1(t)=e^t[/tex]
2nd order Derivative:
[tex]y''_1(t)=e^t[/tex]
Put Them in equation y''-y=0
e^t-e^t=0
0=0
Hence [tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.
For 2nd Solution:
[tex]y_2(t)=cosht[/tex]
In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.
[tex]y_2(t)=cosht[/tex]
First order derivative:
[tex]y'_2(t)=sinht[/tex]
2nd order Derivative:
[tex]y''_2(t)=cosht[/tex]
Put Them in equation y''-y=0
cosht-cosht=0
0=0
Hence [tex]y_2(t)=cosht[/tex] is the solution of equation y''-y=0.
The functions [tex]y(t)=e^t[/tex] and [tex]y(t)=cosht[/tex] are solutions of given differential equation.
We have to prove that given function is a solution of the differential equation.
Given function, [tex]y(t)=e^t[/tex]
Given differential equation are, [tex]y''-y=0[/tex]
So that, [tex]y'(t)=e^t , y''(t)=e^t[/tex]
Substitute values in given differential equation.
[tex]e^t - e^t=0[/tex]
Thus, function [tex]y(t)=e^t[/tex] is solution of given differential equation.
Another function is, [tex]y(t)=cosht[/tex]
So that, [tex]y'(t)=sinht , y''(t)=cosht[/tex]
Substitute values in given differential equation.
[tex]cosht-cosht=0[/tex]
Thus, function [tex]y(t)=cosht[/tex] is solution of given differential equation.
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