Question

Find the function
`y_(1) (t)` which is the solution of
`64y'' +48y' +8y=0` with initial conditions.
`y_(1) (0)=1, y_(1)' (0)=0`

Find the function
`y_(2) (t)` which is the solution of
`64y'' +48y' +8y=0` with initial conditions.
`y_(2) (0)=0, y_(2)' (0)=1`
Find the Wronskian
`W(t)=W(y_(1),y_(2))`
Remark: You can find W by direct computation and use Abel's theorem as a check. You should find that W is not zero and so
`y_(1)` and
`y_(2)`form a fundamental set of solutions of the above equation.

Answer 1

Characteristic equation:


`64r^2+48r+8=0`

`64\left(r^2+\frac34r+\frac18\right)=0`

`\left(r+\frac12\right)\left(r+\frac14\right)=0\implies r=-\frac12,r=-\frac14`

This gives a general solution of


`y_1=C_1e^(-1/2t)+C_2e^(-1/4t)`

With the given initial conditions, you get


`y_1(0)=1\implies 1=C_1+C_2`

`{y_1}'(0)=0\implies 0=-\frac12C_1-\frac14C_2`

`\implies C_1=-1,C_2=2`

so that the first particular solution is


`y_1(t)=-e^(-1/2t)+2e^(-1/4t)`

With the second set of initial conditions, you would have


`y_2(0)=0\implies 0=C_1+C_2`

`{y_2}'(0)=1\implies 1=-\frac12C_1-\frac14C_2`

`\implies C_1=-4,C_2=4`

so that the second particular solution is


`y_2(t)=-4e^(-1/2t)+4e^(-1/4t)`

The Wronskian of the two solution is


`W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=\begin{vmatrix}-e^(-1/2t)+2e^(-1/4t)&-4e^(-1/2t)+4e^(-1/4t)\\-\frac12e^(-1/2t)-\frac12e^(-1/4t)&2e^(-1/2t)-e^(-1/4t)\end{vmatrix}`

`W(y_1,y_2)=e^(-3/4t)`