Respuesta :
Answer:
Therefore the auxiliary solution is [tex]y=A e^{5x}+Be^{-4x}[/tex]
Therefore [tex]e^{-4x} and \ e^{5x}[/tex] are linearly independent
Step-by-step explanation:
Given, the differential equation is
y"-y'-20 y=0
Let [tex]y=e^{mx}[/tex] be the solution of the above differential equation.
y'= [tex]me^{mx}[/tex] and [tex]y"= m^2e^{mx}[/tex]
Then the above differential equation becomes
[tex]m^2e^{mx}-me^{mx}-20 e^{mx}=0[/tex]
[tex]\Rightarrow e^{mx}(m^2-m-20)=0[/tex]
[tex]\Rightarrow (m^2-m-20)=0[/tex]
[tex]\Rightarrow m^2-5m+4m-20=0[/tex]
[tex]\Rightarrow m(m-5) +4(m-5)=0[/tex]
[tex]\Rightarrow (m-5)(m+4)=0[/tex]
[tex]\Rightarrow m=5,-4[/tex]
If two roots of m are real and distinct then the auxiliary solution is
[tex]y=Ae^{ax}+Be^{bx}[/tex] [where a and b are two roots of m]
Therefore the auxiliary solution is [tex]y=A e^{5x}+Be^{-4x}[/tex]
Wronskian
[tex]W(e^{-4x},e^{5x})=\left[\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right][/tex]
[tex]=5e^{-4x}e^{5x}-e^{5x}(-4e^{-4x})[/tex]
[tex]=9e^x[/tex]≠0
Therefore [tex]e^{-4x} and \ e^{5x}[/tex] are linearly independent.[ ∵W≠0]
