Find constants a and b such that the function y = a sin (x) + b cos (x) satisfies the differential equation y'' + y' - 7y = sin (x).

The first thing we must do in this case is find the derivatives:

y = a sin (x) + b cos (x)

y ' = a cos (x) - b sin (x)

y '' = - a sin (x) - b cos (x)

Substituting the values:

(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)

We rewrite:

(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)

sin (x) * ( - a-b-7a) + cos (x) * ( - b + a-7b) = sin (x)

sin (x) * ( - b-8a) + cos (x) * (a-8b) = sin (x)

From here we get the system:

-b-8a = 1

a-8b = 0

Whose solution is:

a = - 8 / 65

b = - 1 / 65

Answer:

constants a and b are:

a = - 8 / 65

b = - 1 / 65