Question

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solutions of the differential equation on the interval (−[infinity], [infinity]). The functions satisfy the differential equation and are linearly independent since the Wronskian W e−4x, e5x =

Answer 1

Answer:

Therefore the auxiliary solution is $y=A e^{5x}+Be^{-4x}$

Therefore $e^{-4x} and \ e^{5x}$ are linearly independent

Step-by-step explanation:

Given, the differential equation is

y"-y'-20 y=0

Let $y=e^{mx}$ be the solution of the above differential equation.

y'= $me^{mx}$    and  $y"= m^2e^{mx}$

Then the above differential equation becomes

$m^2e^{mx}-me^{mx}-20 e^{mx}=0$

$\Rightarrow e^{mx}(m^2-m-20)=0$

$\Rightarrow (m^2-m-20)=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5) +4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=5,-4$

If two roots of m are real and distinct then the auxiliary solution is

$y=Ae^{ax}+Be^{bx}$      [where a and b are two roots of m]

Therefore the auxiliary solution is $y=A e^{5x}+Be^{-4x}$

Wronskian

$W(e^{-4x},e^{5x})=\left[\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right]$

                  $=5e^{-4x}e^{5x}-e^{5x}(-4e^{-4x})$

                  $=9e^x$≠0

Therefore $e^{-4x} and \ e^{5x}$ are linearly independent.[ ∵W≠0]