The general form of a solution of the differential equation is already provided for us:
[tex]y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},[/tex]
where [tex]c_1, c_2 \in \mathbb{R}[/tex]. We now want to find a solution [tex]y[/tex] such that [tex]y(-1)=3[/tex] and [tex]y'(-1)=-3[/tex]. Therefore, all we need to do is find the constants [tex]c_1[/tex] and [tex]c_2[/tex] that satisfy the initial conditions. For the first condition, we have:[tex]y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.[/tex]
For the second condition, we need to find the derivative [tex]y'[/tex] first. In this case, we have:
[tex]y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.[/tex]
Therefore:
[tex]y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.[/tex]
This means that we must solve the following system of equations:
[tex]\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.[/tex]
If we add the equations above, we get:
[tex]\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e} \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.[/tex]
If we now substitute [tex]c_1 = 0[/tex] into either of the equations in the system, we get:
[tex]c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}[/tex]
This means that the solution obeying the initial conditions is:
[tex]\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.[/tex]
Indeed, we can see that:
[tex]y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3[/tex]
[tex]y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,[/tex]
which do correspond to the desired initial conditions.
