Complete Question
The complete question is shown on the first uploaded
Answer:
[tex]y_1[/tex] is not a solution of the differential equation
[tex]y_2[/tex] is not a solution of the differential equation
[tex]y_3[/tex] is not a solution of the differential equation
Step-by-step explanation:
The differential equation given is [tex]y'' + y' = cos2x[/tex]
Let consider the first equation to substitute
[tex]y_1 = cosx +sinx[/tex]
[tex]y_1' = -sinx +cosx[/tex]
[tex]y_1'' = -cosx -sinx[/tex]
So
[tex]y_1'' - y_1' = -cosx -sinx -sinx +cosx[/tex]
[tex]y_1'' + y_1' = -2sinx [/tex]
So
[tex] -2sinx \ne cos2x [/tex]
This means that [tex]y_1[/tex] is not a solution of the differential equation
Let consider the second equation to substitute
[tex]y_2 = cos2x[/tex]
[tex]y_2' = -2sin2x[/tex]
[tex]y_2'' = -4cos2x[/tex]
So
[tex]y_2'' + y_2' = -4cos2x-2sin2x [/tex]
So
[tex] -4cos2x-2sin2x \ne cos2x [/tex]
This means that [tex]y_2[/tex] is not a solution of the differential equation
Let consider the third equation to substitute
[tex]y_3 = sin 2x[/tex]
[tex]y_3' = 2cos 2x[/tex]
[tex]y_3'' = -4sin2x[/tex]
So
[tex]y_3'' + y_3' = -4sin2x - 2cos2x [/tex]
So
[tex] -4sin2x - 2cos2x \ne cos2x [/tex]
This means that [tex]y_3[/tex] is not a solution of the differential equation
