Let T : C∞(R) −→ C∞(R) be the map given by
T(f) = f′′− 2f′− 3f .

Find a linear combination a sin(x) + b cos(x) which solves the equation
f′′− 2f′− 3f = 2 cos(x).

Question

Let T : C∞(R) −→ C∞(R) be the map given by
T(f) = f′′− 2f′− 3f .

Find a linear combination a sin(x) + b cos(x) which solves the equation
f′′− 2f′− 3f = 2 cos(x).

anonymous · Answer

This combination will work
$$f(x)=\frac{1}{5} (-\sin (x)-2 \cos
   (x))
$$
To show that you have to say
$$ f(x) = A \cos(x) + B \sin(x)
$$
and find A and B such that 
f′(x)′− 2f′(x) − 3 f(x) = 2 cos(x).\
Now
$$ f'(x)=B \cos (x)-A \sin (x)\
f''(x)=-A \cos (x)-B \sin (x)\
f''(x)− 2f′(x) − 3 f(x)=-2 ((2 A+B) \cos (x)-(A-2 B)  \sin (x))= 2\cos(x)\
  
$$
Equating the coefficient of cos(x) and sin(x), you get
$$ 2 A+B =-1 \
A- 2 B =0
$$
You solve for A and B
$$A=-\frac{2}{5} \
   B=-\frac{1}{5}
$$
 and you replace.