Question

How do I solve the differential equation `y"-4y'-5y=0` with `y(-1)=3` and `y'(-1)=9`?

Answer 1

As written, there is no solution to the problem.

Definitions: nth-order differential equation: a differential equation in which the highest-order derivative presented is of order n. Thus, a first-order differential equation would involve as its highest-order derivative `y'`.

Ordinary differential equation: a differential equation consisting of a function of one independent variable and its derivatives. An ordinary differential equation might involve `y(x)`, for example, but not `y(x_1, x_2)` in which `x_1` and `x_2` are different independent variables.

For the work shown below, we assume that `y` is a function of `x`, and is thus denoted `y(x)`. However, this is simply a "placeholder" variable so that we can properly define our derivative. If it were instead `y(t)`, then `y' = dy/dt`, for example.

Given the first-order ordinary differential equation above, the first thing we can do is group like terms simplify.
`y - 4y' - 5y = y - 5y - 4y' = -4 (y'+y) = 0`
Thus, dividing by -4 and moving terms to opposite sides:
`y' = -y`.
We will first divide both sides by `y`, and then integrate both sides with respect to `x`
`y' = -y => (y')/y = -1 => int (y')/y dx = int -1 dx => int (y')/ydx = -x + c_1`, where `c_1` is an arbitrary constant.

Recall that `int (du)/u = ln (u)`, and that `e^ln(u) = u...`

`ln (y) = -x + c_1 => e^ln(y) = e^(-x+c_1) = (e^-x)(e^(c_1)) => y = e^(c_1)e^(-x)`. Since our original `c_1` was an arbitrary constant, we can reuse the label, declaring `e^(c_1)` as the "new" `c_1`:

`y(x) = c_1e^-x`

Thus, the initial equation implies a function of the form `y = c_1 e^-x`.

We would normally find the constant `c_1` by using the boundary conditions provided. However, we already have a problem. Since `y' = -c_1e^-x`, `y'(-1)` should be equal to `-y(-1)`. However, this is not the case, as `y'(-1) = 9` while `y(-1)=3`.

Thus, there is no solution to the problem as presented.