I am having some trouble solving this differential equation: $4y''-4y'+2y=0$ when $y(4)=3, y'(4)=3$, if anyone could help, or tell me where I went wrong, I would greatly appreciate it!
I first solved it as a second order Differential Equation and got the quadratic solution $\frac{1}{2}\pm\frac{i}{2}$
Giving the general solution: $y(t)=C_1e^{\frac{t}{2}}cos(\frac{t}{2})+C_2e^{\frac{t}{2}}sin(\frac{t}{2})$
Applying the initial conditions:
$y(4)=C_1e^2cos(2)+C_2e^2sin(2)=3$
$y'(4)=C_1(\frac{1}{2}e^2cos(2)-\frac{1}{2}e^2sin(2)+C_2(\frac{1}{2}e^2sin(2)+\frac{1}{2}e^2cos(2))=3$
And then in order to solve for $C_1$ and $C_2$, I plugged the equations into the following matrix: \begin{bmatrix} e^2cos(2) & e^2sin(2) & 3\\ \frac{1}{2}e^2cos(2)-\frac{1}{2}e^2sin(2) & \frac{1}{2}e^2sin(2)+\frac{1}{2}e^2cos(2) & 3 \end{bmatrix}
And then this is where I start to get confused. Depending on if I use degrees or radians, I will get two different solutions, resulting in these two matrices:
Radians:
\begin{bmatrix}
1 & 0 & -0.538138 \\ 0 & 1 & 0.200222
\end{bmatrix}
Degrees:
\begin{bmatrix}
1 & 0 & 0.391589 \\ 0 & 1 & 0.419928
\end{bmatrix}
Which gives me two potential solutions:
Radians: $y(t)=-0.538138e^{\frac{t}{2}}cos(\frac{t}{2})+0.200222e^{\frac{t}{2}}sin(\frac{t}{2})$
Degrees: $y(t)=0.391589e^{\frac{t}{2}}cos(\frac{t}{2})+0.419928e^{\frac{t}{2}}sin(\frac{t}{2})$
However, neither of these solutions are correct when I plug them into my homework.
The system I am using usually prefers more exact numbers, so I am wondering if there is a way I could solve the equation without matrices to get a more exact value (some combination of $e$ and $cos$ or $sin$ for my $C_1$ and $C_2$ value) or if some part of my process was incorrect.
Again, I would greatly appreciate any help! Thank you so much!