https://file.aiquickdraw.com/mathbot/file/1726588317377gqeidn8p.jpg

The provided image contains a set of linear homogeneous second-order differential equations. Let's solve each of them step-by-step:

### General Approach:
For second-order linear homogeneous differential equations of the form:

\[
y'' + a y' + b y = 0
\]

we can find solutions by solving the **characteristic equation**:

\[
r^2 + a r + b = 0
\]

The nature of the roots \( r_1 \) and \( r_2 \) will determine the general solution of the differential equation:
1. If \( r_1 \) and \( r_2 \) are real and distinct: \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \)
2. If \( r_1 = r_2 \) (real and equal): \( y(t) = (C_1 + C_2 t) e^{r_1 t} \)
3. If \( r_1, r_2 \) are complex conjugates: \( y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \) where \( r_1, r_2 = \alpha \pm i \beta \)

Now let's solve each part.

### a) \( y'' + y' - 6y = 0 \)
1. **Characteristic equation**: \( r^2 + r - 6 = 0 \)
2. Solve using the quadratic formula:
   \[
   r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}
   \]
   Roots are \( r_1 = 2 \) and \( r_2 = -3 \).

3. **General solution**: 
   \[
   y(t) = C_1 e^{2t} + C_2 e^{-3t}
   \]

### b) \( y'' + 6y' + 9y = 0 \)
1. **Characteristic equation**: \( r^2 + 6r + 9 = 0 \)
2. Solve using the quadratic formula:
   \[
   r = \frac{-6 \pm \sqrt{6^2 - 4(1)(9)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 36}}{2} = \frac{-6}{2} = -3
   \]
   Root \( r_1 = r_2 = -3 \) (repeated root).

3. **General solution**: 
   \[
   y(t) = (C_1 + C_2 t) e^{-3t}
   \]

### c) \( y'' - 4y' + 5y = 0 \)
1. **Characteristic equation**: \( r^2 - 4r + 5 = 0 \)
2. Solve using the quadratic formula:
   \[
   r = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2}
   \]
   \( r = 2 \pm i \), which are complex roots.

3. **General solution**: 
   \[
   y(t) = e^{2t} (C_1 \cos(t) + C_2 \sin(t))
   \]

### d) \( y'' + 2y' + 2y = 0 \)
1. **Characteristic equation**: \( r^2 + 2r + 2 = 0 \)
2. Solve using the quadratic formula:
   \[
   r = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2}
   \]
   \( r = -1 \pm i \), which are complex roots.

3. **General solution**: 
   \[
   y(t) = e^{-t} (C_1 \cos(t) + C_2 \sin(t))
   \]

### Summary of Solutions:
- **a)** \( y(t) = C_1 e^{2t} + C_2 e^{-3t} \)
- **b)** \( y(t) = (C_1 + C_2 t) e^{-3t} \)
- **c)** \( y(t) = e^{2t} (C_1 \cos(t) + C_2 \sin(t)) \)
- **d)** \( y(t) = e^{-t} (C_1 \cos(t) + C_2 \sin(t)) \)

Would you like further explanation on any of these? Let me know!

Here are 5 related questions:
1. How do we handle non-homogeneous differential equations?
2. Can we use other methods, like Laplace transforms, to solve these equations?
3. What if the differential equation had more than two terms in derivatives?
4. How do we find the particular solution for non-homogeneous cases?
5. What if the characteristic equation had complex conjugates with different coefficients?

**Tip:** Always check the discriminant \( b^2 - 4ac \) of the characteristic equation to determine the type of roots!

Lösen Sie die folgenden Differentialgleichungen: 
 a) y'' + y' - 6y = 0 
 b) y'' + 6y' + 9y = 0 
 c) y'' - 4y' + 5y = 0 
 d) y'' + 2y' + 2y = 0

Step-by-step solution of second-order linear homogeneous differential equations including characteristic equations, complex roots, and their general solutions. Learn to solve equations like y'' + y' - 6y = 0, y'' + 6y' + 9y = 0, and more.

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