Answer:
Y = [tex]e^{t}[/tex] + [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18
Step-by-step explanation:
y''' − 3y'' + 3y' − y = ex − x + 21
Homogeneous solution:
First we propose a solution:
Yh = [tex]e^{r*t}[/tex]
Y'h = [tex]r*e^{r*t}[/tex]
Y''h = [tex]r^{2}*e^{r*t}[/tex]
Y'''h = [tex]r^{3}*e^{r*t}[/tex]
Now we solve the following equation:
Y'''h - 3*Y''h + 3*Y'h - Yh = 0
[tex]r^{3}*e^{r*t}[/tex] - 3*[tex]r^{2}*e^{r*t}[/tex] + 3*[tex]r*e^{r*t}[/tex] - [tex]e^{r*t}[/tex] = 0
[tex]r^{3} - 3r^{2} + 3r - 1 = 0[/tex]
To solve the equation we must propose a solution to the polynomial
:
r = 1
To find the other r we divide the polynomial by (r-1) as you can see
attached:
solving the equation:
(r-1)([tex]r^{2} - 2r + 1[/tex]) = 0
[tex]r^{2} - 2r + 1[/tex] = 0
r = 1
So we have 3 solution [tex]r_{1} = r_{2} =r_{3}[/tex] = 1
replacing in the main solution
Yh = [tex]e^{t}[/tex] + [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex]
The t and [tex]t^{2}[/tex] is used because we must have 3 solution linearly independent
Particular solution:
We must propose a Yp solution:
Yp = [tex]c_{1} (t^{3} + t^{2} + t + c_{4} )e^{t} + c_{2} t + c_{3}[/tex]
Y'p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}( 3t^{2} + 2t + 1 )e^{t} + c_{2}[/tex]
Y''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}(6t + 2)e^{t}[/tex]
Y'''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + 6c_{1}e^{t}[/tex]
Y'''p - 3*Y''p + 3*Y'p - Yp = [tex]e^{t} - t + 21[/tex]
[tex]6c_{1}e^{t} - 18c_{1} te^{t} - 6c_{1} e^{t} + 6c_{1} te^{t} + 9c_{1} t^{2} e^{t} + 3c_{1}e^{t} + 3c_{2} - c_{2} t - c_{3}[/tex] = [tex]e^{t} - t + 21[/tex]
equalizing coefficients of the same function:
- 12c_{1} = 0
9c_{1} = 0
3c_{1} = 0
c_{1} = 0
3c_{2} - c_{3} = 21 => c_{5} = [tex]\frac{1}{3}[/tex]
-c_{2} = -1
c_{2} = 1
c_{3} = -18
Then we have:
Y = [tex]e^{t}[/tex] + [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18
