To solve the differential equation given in the image:

$y^{(5)} - 6y^{(4)} + 9y''' - 6y'' + 8y' = 0$

we'll proceed as follows:

1. Form the Characteristic Equation: For a linear differential equation of the form $y^{(5)} - 6y^{(4)} + 9y''' - 6y'' + 8y' = 0$, the corresponding characteristic equation is: $r^5 - 6r^4 + 9r^3 - 6r^2 + 8r = 0$

2. Factor Out $r$: We can factor out $r$ from the characteristic equation: $r(r^4 - 6r^3 + 9r^2 - 6r + 8) = 0$ This gives one root, $r = 0$.

3. Solve the Remaining Polynomial $r^4 - 6r^3 + 9r^2 - 6r + 8 = 0$: We can try factoring or using the Rational Root Theorem to find other roots. In this case, we observe that it factors as: $(r - 2)(r - 2)(r - 2)(r + 2) = 0$ or $(r - 2)^3 (r + 2) = 0$

   This gives the roots:

   • $r = 0$
   • $r = 2$ (with multiplicity 3)
   • $r = -2$

4. Write the General Solution: Based on these roots, we can write the general solution for $y(t)$ as follows: $y(t) = c_1 + c_2 e^{2t} + c_3 t e^{2t} + c_4 t^2 e^{2t} + c_5 e^{-2t}$ where $c_1, c_2, c_3, c_4,$ and $c_5$ are arbitrary constants.

Final Answer:

$y(t) = c_1 + c_2 e^{2t} + c_3 t e^{2t} + c_4 t^2 e^{2t} + c_5 e^{-2t}$

Would you like a detailed breakdown of each step or have any questions?

Related Questions

1. How would you find the characteristic equation for a second-order linear differential equation?
2. What is the significance of the multiplicity of roots in the characteristic equation?
3. How does the presence of a zero root affect the solution of a differential equation?
4. How would you solve a differential equation if complex roots were present?
5. What methods can be used to factor higher-degree polynomials?

Tip:

Remember, the multiplicity of a root in the characteristic equation affects the solution's form by introducing polynomial terms in $t$ (e.g., $t$, $t^2$) for each repeated root.