Answer:
(See attached graph)
Explanation:
To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation
where the values of
are the roots:

Since the values of
are complex conjugate roots, then the general solution is
where
.
Thus, the general solution for our given differential equation is
.
To account for both initial conditions, take the derivative of
, thus,
![y'(x)=-2e^(-2x)[C_1cos(√(6)x+C_2sin(√(6)x)]+e^(-2x)[-C_1√(6)sin(√(6)x)+C_2√(6)cos(√(6)x)]](https://img.qammunity.org/2023/formulas/mathematics/college/53opv5ctyadfdi5wx4chzxqkiszm5sfl7q.png)
Now, we can create our system of equations given our initial conditions:
![y(x)=e^(-2x)[C_1cos(√(6)x)+C_2sin(√(6)x)]\\\\y(0)=e^(-2(0))[C_1cos(√(6)(0))+C_2sin(√(6)(0))]=3\\\\C_1=3](https://img.qammunity.org/2023/formulas/mathematics/college/lgr3f24boc6013cv2pwkh4ka9isma0bkx7.png)
![y'(x)=-2e^(-2x)[C_1cos(√(6)x+C_2sin(√(6)x)]+e^(-2x)[-C_1√(6)sin(√(6)x)+C_2√(6)cos(√(6)x)]\\\\y'(0)=-2e^(-2(0))[C_1cos(√(6)(0))+C_2sin(√(6)(0))]+e^(-2(0))[-C_1√(6)sin(√(6)(0))+C_2√(6)cos(√(6)(0))]=-2\\\\-2C_1+√(6)C_2=-2](https://img.qammunity.org/2023/formulas/mathematics/college/5759xy0pni97v8davombsezx9ndg3q1pj4.png)
We then solve the system of equations, which becomes easy since we already know that
:

Thus, our final solution is:
![y(x)=e^(-2x)[C_1cos(√(6)x)+C_2sin(√(6)x)]\\\\y(x)=e^(-2x)[3cos(√(6)x)+(2√(6))/(3)sin(√(6)x)]](https://img.qammunity.org/2023/formulas/mathematics/college/yu5t3z0ogeo0npt2va5xh8u8sl0uuudzse.png)