Question

Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!! +y′= cos 2x;y"= cosx+sinx,y # = cos 2x,y$ =sin 2x

Answer 1

Complete Question

The complete question is shown on the first uploaded

Answer:

$y_1$ is not a solution of the differential equation

 $y_2$ is not a solution of the differential equation

$y_3$ is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is $y'' + y' = cos2x$

Let consider the first equation to substitute

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

So

$y_1'' - y_1' = -cosx -sinx -sinx +cosx$

$y_1'' + y_1' = -2sinx$

So

$-2sinx \ne cos2x$

This means that $y_1$ is not a solution of the differential equation

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

So

$y_2'' + y_2' = -4cos2x-2sin2x$

So

$-4cos2x-2sin2x \ne cos2x$

This means that $y_2$ is not a solution of the differential equation

Let consider the third equation to substitute

$y_3 = sin 2x$

   $y_3' = 2cos 2x$

    $y_3'' = -4sin2x$

So

$y_3'' + y_3' = -4sin2x - 2cos2x$

So

$-4sin2x - 2cos2x \ne cos2x$

This means  that  $y_3$ is not a solution of the differential equation