Question

In Problems 22–25, use the method described in Problem 21 to find a general solution to the given equation.

23. 5Y' + 4Y = 0

Short answer

The general solution is $y\left(t\right)=c{e}^{-\frac{4}{5}t}$.

Step-by-step solution

Substitute y = ert

The given differential equation is 5y' + 4y = 0

Substitute y = e^rt and y' = re^rt,

$\begin{array}{rcl}5\left(r{e}^{rt}\right)+4\left(e^{rt}\right)& =& 0\\ (5r+4)\left(e^{rt}\right)& =& 0\\ 5r+4& =& 0\end{array}$

Therefore,

$r=-\frac{4}{5}$

Find the general solution.

Thus, the general solution is $y\left(t\right)=c{e}^{-\frac{4}{5}t}$.