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The differential equation provided is:

\[
y'' - 5y' + 5y = 0
\]

This is a second-order linear homogeneous differential equation with constant coefficients. To solve this, we first find the characteristic equation by assuming a solution of the form \( y(t) = e^{rt} \). Substituting this into the differential equation gives:

\[
r^2 e^{rt} - 5r e^{rt} + 5e^{rt} = 0
\]

Factoring out \( e^{rt} \) (which is never zero), we are left with the characteristic equation:

\[
r^2 - 5r + 5 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
r = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(5)}}{2(1)} = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}
\]

The roots of the characteristic equation are:

\[
r_1 = \frac{5 + \sqrt{5}}{2}, \quad r_2 = \frac{5 - \sqrt{5}}{2}
\]

Since the roots are real and distinct, the general solution to the differential equation is:

\[
y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}
\]

Substitute \( r_1 \) and \( r_2 \) back into the solution:

\[
y(t) = c_1 e^{\frac{5 + \sqrt{5}}{2} t} + c_2 e^{\frac{5 - \sqrt{5}}{2} t}
\]

This is the general solution to the given differential equation.

Would you like further details or have any questions?

Here are 5 related questions you may consider:
1. How do we solve characteristic equations with complex roots?
2. How does the nature of the roots affect the form of the solution?
3. What happens if the coefficients of the differential equation are not constant?
4. Can this method be used to solve non-homogeneous differential equations?
5. How does this approach extend to higher-order differential equations?

**Tip**: Always check the discriminant of the characteristic equation to determine the type of solution (real, repeated, or complex roots).

Find the general solution of the given differential equation: y'' - 5y' + 5y = 0

Find the general solution of the second-order linear homogeneous differential equation y'' - 5y' + 5y = 0. This example uses the characteristic equation to solve for real and distinct roots, yielding the general solution y(t) = c1 e^((5 + √5)/2 t) + c2 e^((5 - √5)/2 t).

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