To determine which ODE corresponds to the given general solution, let's first analyze the general solution:

Given General Solution:

$y = c_1 e^{2x} + c_2 e^{-5x}$

Characteristic Equation:

For a second-order linear homogeneous differential equation of the form: $y'' + by' + cy = 0$ the solution is based on the roots of the characteristic equation: $r^2 + br + c = 0$

For the given general solution, the roots are $r = 2$ and $r = -5$.

So, the characteristic equation corresponding to these roots would be: $(r - 2)(r + 5) = 0$ Expanding this: $r^2 + 3r - 10 = 0$

This gives us the corresponding differential equation: $y'' + 3y' - 10y = 0$

Matching the Correct ODE:

Now, let's match this to the given options:

1. $y'' + 6y' + 10y = 0$

   • Characteristic equation: $r^2 + 6r + 10 = 0$.
   • Roots: $r = -3 \pm i$, not matching the given roots.

2. $y'' + 6y' + 12y = 0$

   • Characteristic equation: $r^2 + 6r + 12 = 0$.
   • Roots: $r = -3 \pm i\sqrt{3}$, not matching the given roots.

3. $y'' + 6y' + 15y = 0$

   • Characteristic equation: $r^2 + 6r + 15 = 0$.
   • Roots: $r = -3 \pm 2i$, not matching the given roots.

4. $y'' + 6y' + 8y = 0$

   • Characteristic equation: $r^2 + 6r + 8 = 0$.
   • Roots: $r = -2 \pm -4$, not matching the given roots.

Conclusion:

The correct ODE from the provided options is:

Would you like further elaboration on how to solve these types of equations or other aspects of the problem?

Relative Questions:

1. How do you find the characteristic equation of a differential equation?
2. What are the methods to solve second-order linear homogeneous differential equations?
3. How do initial conditions affect the solution of a second-order ODE?
4. Can complex roots occur in the characteristic equation? What would the general solution look like then?
5. What are the real-world applications of second-order differential equations?

Tip:

Always check the roots of the characteristic equation to ensure they match the exponents in the general solution for homogeneous linear differential equations.