Respuesta :
Answer:
The complete solution is
[tex]\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43[/tex]
Step-by-step explanation:
Given differential equation is
3y"- 8y' - 3y =4
The trial solution is
[tex]y = e^{mx}[/tex]
Differentiating with respect to x
[tex]y'= me^{mx}[/tex]
Again differentiating with respect to x
[tex]y''= m ^2 e^{mx}[/tex]
Putting the value of y, y' and y'' in left side of the differential equation
[tex]3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0[/tex]
[tex]\Rightarrow 3m^2-8m-3=0[/tex]
The auxiliary equation is
[tex]3m^2-8m-3=0[/tex]
[tex]\Rightarrow 3m^2 -9m+m-3m=0[/tex]
[tex]\Rightarrow 3m(m-3)+1(m-3)=0[/tex]
[tex]\Rightarrow (3m+1)(m-3)=0[/tex]
[tex]\Rightarrow m = 3, -\frac13[/tex]
The complementary function is
[tex]y= Ae^{3x}+Be^{-\frac13 x}[/tex]
y''= D², y' = D
The given differential equation is
(3D²-8D-3D)y =4
⇒(3D+1)(D-3)y =4
Since the linear operation is
L(D) ≡ (3D+1)(D-3)
For particular integral
[tex]y_p=\frac 1{(3D+1)(D-3)} .4[/tex]
[tex]=4.\frac 1{(3D+1)(D-3)} .e^{0.x}[/tex] [since [tex]e^{0.x}=1[/tex]]
[tex]=4\frac{1}{(3.0+1)(0-3)}[/tex] [ replace D by 0 , since L(0)≠0]
[tex]=-\frac43[/tex]
The complete solution is
y= C.F+P.I
[tex]\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43[/tex]
