Question

Solve the following differential equation:
y'' - 2y' + 5y = 0

Answer 1

To solve the differential equation \( y'' - 2y' + 5y = 0 \), we start by finding the characteristic equation associated with the homogeneous equation. The characteristic equation is obtained by substituting \( y = e^{rt} \), where \( r \) is a constant, into the differential equation.

1. Substitute \( y = e^{rt} \):
\[
y' = re^{rt}, \quad y'' = r^2 e^{rt}
\]

2. Plug these into the differential equation:
\[
r^2 e^{rt} - 2re^{rt} + 5e^{rt} = 0
\]

3. Factor out \( e^{rt} \) (which is never zero):
\[
e^{rt}(r^2 - 2r + 5) = 0
\]

4. Thus, we need to solve the characteristic equation:
\[
r^2 - 2r + 5 = 0
\]

5. Use the quadratic formula to find the roots:
\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2}
\]

Simplifying this gives:
\[
r = \frac{2 \pm 4i}{2} = 1 \pm 2i
\]

6. The roots of the characteristic equation are \( r = 1 + 2i \) and \( r = 1 - 2i \). These are complex conjugates.

7. The general solution to the differential equation can be written using the standard form for complex roots:
\[
y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t))
\]
where \( \alpha = 1 \) and \( \beta = 2 \).

Thus, the general solution is:
\[
y(t) = e^{t} (C_1 \cos(2t) + C_2 \sin(2t))
\]

where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.