Let's walk through the steps shown in the images:

First Image:

The differential equation provided is:

$y'' - 6y' + 13y = 0$

This is a second-order linear homogeneous differential equation with constant coefficients.

Step 1: Write the characteristic equation

$m^2 - 6m + 13 = 0$

This is obtained by assuming a solution of the form $y = e^{mx}$.

Step 2: Solve the characteristic equation using the quadratic formula:

$m = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(13)}}{2(1)}$

This simplifies to: $m = \frac{6 \pm \sqrt{36 - 52}}{2}$ $m = \frac{6 \pm \sqrt{-16}}{2}$ $m = \frac{6 \pm 4i}{2}$ $m = 3 \pm 2i$

So, the roots are complex: $m_1 = 3 + 2i$ and $m_2 = 3 - 2i$.

Step 3: General solution of the differential equation

Since the roots are complex, the general solution is of the form: $y(x) = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x))$ Where $\alpha = 3$ and $\beta = 2$.

Thus, the general solution becomes: $y(x) = e^{3x} (C_1 \cos(2x) + C_2 \sin(2x))$

Second Image:

It seems to contain the same result: $y(x) = e^{3x} (C_1 \cos(2x) + C_2 \sin(2x))$ This matches the correct solution form.

Conclusion:

Yes, the solution provided in the notebook is correct. The process of solving the characteristic equation and forming the general solution based on the complex roots is accurate.

Would you like a more detailed explanation of any part of this solution, or do you have further questions?

5 Related Questions:

1. What happens if the roots of the characteristic equation are real and distinct?
2. How would the solution change if the characteristic equation had repeated roots?
3. What are the conditions for a homogeneous differential equation to have complex roots?
4. Can you solve a similar equation with different coefficients, such as $y'' + 2y' + 5y = 0$?
5. How do we apply initial conditions to find the constants $C_1$ and $C_2$?

Tip:

When solving a second-order differential equation, always ensure the roots of the characteristic equation are interpreted correctly (real, repeated, or complex), as this determines the form of the general solution.