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

Mar 12, 2015

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 $y \left(x\right)$, for example, but not $y \left({x}_{1} , {x}_{2}\right)$ 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 $y \left(x\right)$. However, this is simply a "placeholder" variable so that we can properly define our derivative. If it were instead $y \left(t\right)$, then $y ' = \frac{\mathrm{dy}}{\mathrm{dt}}$, for example.

Given the first-order ordinary differential equation above, the first thing we can do is group like terms simplify.
$y - 4 y ' - 5 y = y - 5 y - 4 y ' = - 4 \left(y ' + y\right) = 0$
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 $x$
$y ' = - y \implies \frac{y '}{y} = - 1 \implies \int \frac{y '}{y} \mathrm{dx} = \int - 1 \mathrm{dx} \implies \int \frac{y '}{y} \mathrm{dx} = - x + {c}_{1}$, where ${c}_{1}$ is an arbitrary constant.

Recall that $\int \frac{\mathrm{du}}{u} = \ln \left(u\right)$, and that ${e}^{\ln} \left(u\right) = u \ldots$

$\ln \left(y\right) = - x + {c}_{1} \implies {e}^{\ln} \left(y\right) = {e}^{- x + {c}_{1}} = \left({e}^{-} x\right) \left({e}^{{c}_{1}}\right) \implies y = {e}^{{c}_{1}} {e}^{- x}$. Since our original ${c}_{1}$ was an arbitrary constant, we can reuse the label, declaring ${e}^{{c}_{1}}$ as the "new" ${c}_{1}$:

$y \left(x\right) = {c}_{1} {e}^{-} x$

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 $y ' = - {c}_{1} {e}^{-} x$, $y ' \left(- 1\right)$ should be equal to $- y \left(- 1\right)$. However, this is not the case, as $y ' \left(- 1\right) = 9$ while $y \left(- 1\right) = 3$.

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