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

Therefore the auxiliary solution is
`y=A e^(5x)+Be^(-4x)`

Therefore
`e^(-4x) and \ e^(5x)` are linearly independent

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`

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]