Respuesta :
Answer:
Therefore the auxiliary solution is
[tex]y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t][/tex]
Step-by-step explanation:
To find auxiliary equation we have to put [tex]u=e^{mt}[/tex] in the given differential equation.
The degree of the differential equation is 3.
Therefore the number of root of the differential equation is 3.
Let [tex]\lambda_1[/tex], [tex]\lambda_2[/tex] and [tex]\lambda_3[/tex] be three roots of the auxiliary equation.
- If three roots are real and equal.
Then [tex]y= e^{\lambda_1t} (C_1+C_2t+C_3t^2)[/tex]
- If three roots are real and distinct.
Then [tex]y=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}+C_3e^{\lambda_3x}[/tex]
- If two roots imaginary and one root real , [tex]\lambda_1= a+ib \ and \ \lambda_2= a-ib[/tex]
Then [tex]y=C_1e^{\lambda_3t}+e^{at}(C_2sin \ bt+C_3cos \ bt)[/tex]
Now, [tex]u=e^{mt}[/tex],[tex]u'=me^{mt},u"=m^2e^{mt} \ and \ u'"= m^3e^{mt}[/tex]
Given differential equation is
[tex]\frac{d^3u}{dt^3} +\frac{d^2u}{dt^2}-2u=0[/tex]
The auxiliary equation is
[tex](m^3+m^2-2)e^{mt}=0[/tex]
[tex]\Rightarrow (m^3+m^2-2)=0[/tex]
[tex]\Rightarrow m^3-m^2+2m^2-2m+2m-2=0[/tex]
[tex]\Rightarrow m^2(m-1)+2m(m-1)+2(m-1)=0[/tex]
[tex]\Rightarrow (m-1)(m^2+2m+2)=0[/tex]
[tex]\Rightarrow m= 1,\frac{-2\pm\sqrt{2^2-4.2.1} }{2.1}[/tex]
[tex]\Rightarrow m= 1,\frac{-2\pm\sqrt{-4} }{2.1}[/tex]
[tex]\Rightarrow m= 1,-1\pm i[/tex]
Here [tex]\lambda_1= -1+i , \lambda_2=- 1-i \ and \ \lambda_3=1[/tex]
Therefore ,
[tex]y=C_1e^{1.t}+e^{-1.t}[C_2sin \ (1.t)+C_3 cos \ (1.t)][/tex]
[tex]\Rightarrow y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t][/tex]
