Question

align="absmiddle" class="latex-formula"> 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:

$64r^2+48r+8=0$
$64\left(r^2+\dfrac34r+\dfrac18\right)=0$
$\left(r+\dfrac12\right)\left(r+\dfrac14\right)=0\implies r=-\dfrac12,r=-\dfrac14$

This gives a general solution of

$y_1=C_1e^{-1/2t}+C_2e^{-1/4t}$

With the given initial conditions, you get

$y_1(0)=1\implies 1=C_1+C_2$
${y_1}'(0)=0\implies 0=-\dfrac12C_1-\dfrac14C_2$
$\implies C_1=-1,C_2=2$

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

$y_2(0)=0\implies 0=C_1+C_2$
${y_2}'(0)=1\implies 1=-\dfrac12C_1-\dfrac14C_2$
$\implies C_1=-4,C_2=4$

so that the second particular solution is

$y_2(t)=-4e^{-1/2t}+4e^{-1/4t}$

The Wronskian of the two solution is

$W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=\begin{vmatrix}-e^{-1/2t}+2e^{-1/4t}&-4e^{-1/2t}+4e^{-1/4t}\\-\frac12e^{-1/2t}-\frac12e^{-1/4t}&2e^{-1/2t}-e^{-1/4t}\end{vmatrix}$
$W(y_1,y_2)=e^{-3/4t}$