Question

y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the second order initial value problem with initial conditions y(−1)=3,y′(−1)=−3

Answer 1

The general form of a solution of the differential equation is already provided for us:

`y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^(-x),`

where
`c_1, c_2 \in \mathbb{R}`. We now want to find a solution
`y` such that
`y(-1)=3` and
`y'(-1)=-3`. Therefore, all we need to do is find the constants
`c_1` and
`c_2` that satisfy the initial conditions. For the first condition, we have:
`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.`

For the second condition, we need to find the derivative
`y'` first. In this case, we have:

`y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^(-x)\right)' = c_1\textrm{e}^x - c_2\textrm{e}^(-x).`

Therefore:

`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.`

This means that we must solve the following system of equations:

`\begin{cases}c_1\textrm{e}^(-1) + c_2\textrm{e} = 3 \\ c_1\textrm{e}^(-1) - c_2\textrm{e} = -3\end{cases}.`

If we add the equations above, we get:

`\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.`

If we now substitute
`c_1 = 0` into either of the equations in the system, we get:

`c_2 \textrm{e} = 3 \iff c_2 = \frac{3}{\textrm{e}} = 3\textrm{e}^(-1.)`

This means that the solution obeying the initial conditions is:

`\boxed{y(x) = 3\textrm{e}^(-1) * \textrm{e}^(-x) = 3\textrm{e}^(-x-1)}.`

Indeed, we can see that:

`y(-1) = 3\textrm{e}^(-(-1) -1) = 3\textrm{e}^(1-1) = 3\textrm{e}^0 = 3`

`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,`

which do correspond to the desired initial conditions.