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:

$c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}$

This means that the solution obeying the initial conditions is:

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

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.