Let T : C∞(R) −→ C∞(R) be the map given by
T (f ) = f ′′ − 2f ′ − 3f
Find a solution to f′′ −2f′ −3f =2cos(x) which also satisfies f(0) = 2 and f′(0) = 3.

Question

Let T : C∞(R) −→ C∞(R) be the map given by
T (f ) = f ′′ − 2f ′ − 3f
Find a solution to f′′ −2f′ −3f =2cos(x) which also satisfies f(0) = 2 and f′(0) = 3.

anonymous · Answer

$$\frac{d^2y}{dx^2}-2\frac{dy}{dx}-3y=2\cos(x)\implies y=c_1xe^{-x}+c_2e^{3x}-\frac{\sin(x)}{5}-\frac{2\cos(x)}{5}$$$$\implies y=\frac{1}{5}(5e^{-x}+7e^{3x}-\sin(x)-2\cos(x))$$

anonymous · Answer

how did you get 1/5 sin(x) - 2/5 cos(x)? Its the linear combination solving the transformation?