Respuesta :
Answer:
The general solution is
[tex]y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))[/tex]
+ [tex]\frac{x^2}{2}[/tex]
Step-by-step explanation:
Step :1:-
Given differential equation y(4) − 2y''' + y'' = e^x + 1
The differential operator form of the given differential equation
[tex](D^4 -2D^3+D^2)y = e^x+1[/tex]
comparing f(D)y = e^ x+1
The auxiliary equation (A.E) f(m) = 0
[tex]m^4 -2m^3+m^2 = 0[/tex]
[tex]m^2(m^2 -2m+1) = 0[/tex]
[tex](m^2 -2m+1) this is the expansion of (a-b)^2[/tex]
[tex]m^2 =0 and (m-1)^2 =0[/tex]
The roots are m=0,0 and m =1,1
complementary function is [tex]y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x[/tex]
Step 2:-
The particular equation is [tex]\frac{1}{f(D)} Q[/tex]
P.I = [tex]\frac{1}{D^2(D-1)^2} e^x+1[/tex]
P.I = [tex]\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}[/tex]
P.I = [tex]I_{1} +I_{2}[/tex]
[tex]\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}[/tex]
[tex]\frac{1}{D} means integration[/tex]
[tex]\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx[/tex]
applying in integration u v formula
[tex]\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx[/tex]
[tex]I_{1}[/tex] = [tex]\frac{1}{D^2(D-1)^2} e^x[/tex]
[tex]\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))[/tex]
[tex]\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))[/tex]
[tex]I_{2}[/tex][tex]= \frac{1}{D^2(D-1)^2}e^{0x}[/tex]
[tex]\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x[/tex]
again integration [tex]\frac{1}{D} x = \frac{x^2}{2!}[/tex]
The general solution is [tex]y = y_{C} +y_{P}[/tex]
[tex]y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))[/tex]
+ [tex]\frac{x^2}{2!}[/tex]
