Question

Section 5.2 Problem 9:

Find the general solution.

`y'' - 2y' + 3y = 0`
​

Answer 1

`y(x)=C_1e^(-x)+C_2e^(3x)`

Explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation
`am^2+bm+c=0` where the values of
`m` are the roots:

`y''-2y'+3y=0\\\\m^2-2m+3=0\\\\(m+1)(m-3)=0\\\\m=-1,\:m=3`

Since the values of
`m` are distinct real roots, then the general solution is
`y(x)=C_1e^(m_1x)+C_2e^(m_2x)`.

Thus, the general solution for our given differential equation is
`y(x)=C_1e^(-x)+C_2e^(3x)`.