Solution :
Given :
2y''' + 15y'' + 24y' + 11y= 0
Let x = independent variable
is a differential equation.
If
![Q(x) \\eq 0](https://img.qammunity.org/2022/formulas/mathematics/college/pfqh80dwuxh9mc1koc32ltsg8a27vecsyh.png)
It is non homogeneous then,
The general solution = complementary solution + particular integral
If Q(x) = 0
It is called the homogeneous then the general solution = complementary solution.
2y''' + 15y'' + 24y' + 11y= 0
![$(2D^3+15D^2+24D+11)y=0$](https://img.qammunity.org/2022/formulas/mathematics/college/icsrq5kj86ohoxjsg665aij8xe4lvrjccq.png)
Auxiliary equation,
![$2m^3+15m^2+24m +11 = 0$](https://img.qammunity.org/2022/formulas/mathematics/college/j78cwfmcb26f8df9hhi3r2tzpe3200f8as.png)
-1 | 2 15 24 11
| 0 -2 - 13 -11
2 13 11 0
∴
![2m^2+13m+11=0](https://img.qammunity.org/2022/formulas/mathematics/college/9ardui5cxz1d02d2a37ckk1rmel2n008rg.png)
The roots are
![$=(-b\pm √(b^2-4ac))/(2a)$](https://img.qammunity.org/2022/formulas/mathematics/college/801tnibm3ju91axz8s50ag2oqzca60txmy.png)
![$=(-13\pm √(13^2-4(11)(2)))/(2(2))$](https://img.qammunity.org/2022/formulas/mathematics/college/drdehj79kd3zxyhzhiqymukwjh45wefc7b.png)
![$=(-13\pm9)/(4)$](https://img.qammunity.org/2022/formulas/mathematics/college/rlg3t0hyp48twks7khk4p6q1q5tacbvp9i.png)
![$=-5.5, -1$](https://img.qammunity.org/2022/formulas/mathematics/college/zn6y70hxyzh76iw4toek5x3wojbswo4asw.png)
So,
![m_1, m_2, m_3 = -1, -1, -5.5](https://img.qammunity.org/2022/formulas/mathematics/college/48tjfypjhjv7sk73oic6nfyueptbouajbd.png)
Then the general solution is :
![$= (c_1+c_2 x)e^(-x) + c_3 \ e^(-5.5x)$](https://img.qammunity.org/2022/formulas/mathematics/college/5x22d7eb08bfar0vrifvfhy284awti1bif.png)