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The given differential equation is:

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

This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we follow these steps:

### Step 1: Find the characteristic equation
We assume a solution of the form \( y(t) = e^{rt} \), where \(r\) is a constant. Substituting \( y(t) = e^{rt} \) into the differential equation:

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

Factoring out \( e^{rt} \) (which is never zero):

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

This simplifies to the characteristic equation:

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

### Step 2: Solve the characteristic equation
We solve the quadratic equation \( r^2 - 5r + 5 = 0 \) using the quadratic formula:

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

\[
r = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}
\]

So the roots are:

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

### Step 3: Write the general solution
Since we have two distinct real roots, the general solution to the differential equation is:

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

Substituting the values of \( r_1 \) and \( r_2 \):

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

This is the general solution to the differential equation.

---

Would you like any additional details on how this was solved or how the quadratic formula was applied? 

Here are 5 related questions to further explore:

1. What happens if the characteristic equation has complex roots?
2. How would you solve a non-homogeneous second-order differential equation?
3. What is the solution method when the roots of the characteristic equation are repeated?
4. How can you solve second-order differential equations using Laplace transforms?
5. How does the solution behavior change when coefficients in the differential equation are functions of \(t\)?

**Tip:** When solving a second-order differential equation, always check if the characteristic equation has real, complex, or repeated roots, as the form of the general solution will vary accordingly.

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

Learn how to solve the second-order homogeneous differential equation y'' - 5y' + 5y = 0. This solution involves finding the characteristic equation and applying the quadratic formula. Suitable for college-level students in calculus or differential equations.

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