Question

Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying the initial conditions y(0) = – 7, y'(0) = 23 y = ( Preview

Answer 1

a.
`w(t)=-12e^(2t)`

b.
`y(t)=-(9)/(2)e^(7t)-(5)/(2)e^(-5t)`

Explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

`(D^2-2D-35)=0`

By factorization method we are finding the solution

`D^2-7D+5D-35=0`

`(D-7)(D+5)=0`

Substitute each factor equal to zero

D-7=0 and D+5=0

D=7 and D=-5

Therefore ,

General solution is

`y(x)=C_1e^(7t)+C_2e^(-5t)`

Let
`y_1=e^(7t) \;and \;y_2=e^(-5t)`

We have to find Wronskian

`w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}`

Substitute values then we get

`w(t)=\begin{vmatrix}e^(7t)&e^(-5t)\\7e^(7t)&-5e^(-5t)\end{vmatrix}`

`w(t)=-5e^(7t)\cdot e^(-5t)-7e^(7t)\cdot e^(-5t)=-5e^(7t-5t)-7e^[7t-5t}`

`w(t)=-5e^(2t)-7e^(2t)=-12e^(2t)`

a.
`w(t)=-12e^(2t)`

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

`y(0)=C_1+C_2`

`C_1+C_2=-7`....(equation I)

`y'(t)=7C_1e^(7t)-5C_2e^(-5t)`

`y'(0)=7C_1-5C_2`

`7C_1-5C_2=23`......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminate
`C_1`

Then we get
`C_2=-(5)/(2)`

Substitute the value of
`C_2` in I equation then we get

`C_1-(5)/(2)=-7`

`C_1=-7+(5)/(2)=(-14+5)/(2)=-(9)/(2)`

Hence, the general solution is

b.
`y(t)=-(9)/(2)e^(7t)-(5)/(2)e^(-5t)`