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Can you please explain to me, what the answer to the following question is, I do not know any way how to proceed:

$$\ddot{x}_1 = -5x_1+4x_2, \quad x_1(0)=2, \quad \dot{x}_1(0)=-1$$ $$\ddot{x}_2 = 4x_1-5x_2, \quad x_2(0)=0, \quad \dot{x}_2(0)=1$$

If $t \rightarrow x(t)$ solves the above initial value problem, which initial value problem solves $y(t) := T^{-1} \ x(t)$ ?

I have $T:= \left(\begin{matrix}1 & 1 \\1 & -1 \\\end{matrix}\right)$, so $T^{-1}:= \left(\begin{matrix}\frac 1 2 & \frac 1 2 \\\frac 1 2 & -\frac{1}2 \\\end{matrix}\right)$.

My decoupled equations are: $$\left(\begin{matrix}\ddot{y}_1 \\\ddot{y}_2 \\\end{matrix}\right) = \left(\begin{matrix}-1 & 0 \\0 & -9 \\\end{matrix}\right) \left(\begin{matrix}y_1 \\y_2 \\\end{matrix}\right)$$ Then I solved the decoupled differential equations: $$y_1(t)=C_1cos(t)+C_2sin(t)$$ $$y_2(t)=C_1cos(3t)+C_2sin(3t)$$

If its relevant, I used the following eqautions to decouple the system: $$\ddot{x} = Ax, \ \mathrm{where} \ A:= \left(\begin{matrix}-5 & 4 \\4 & -5 \\\end{matrix}\right)$$$$\ddot{y} = Dy, \ \mathrm{where} \ D:= \left(\begin{matrix}-1 & 0 \\0 & -9 \\\end{matrix}\right)$$

And now I gues I somehow need to "retransform" (with $y=T^{-1}x$ ?) my transformed solution for $y$ in order to find the solution for the whole initial value problem. But I do not know how to do that exactly. Apart from that I would be great if you could explain to me the answer of the bold question.

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You had solved in fact your system, without finding initial conditions.

Let us rewrite the solution in a simpler way because there is not need to consider the diagonalized form.

First of all:

$Y=T^{-1}X \ \iff \ X=TY$ which implies that

$$\tag{1}\ddot X = T \ddot Y \ \ \ \text{because} \ T \ \text{is a constant matrix.}$$

Let us consider the differential system under the form $\ddot X=A X$, then, using (1)

$$\tag{2}(T \ddot Y) = A (T Y) \ \iff \ \ddot Y=\underbrace{(T^{-1}AT)}_D Y $$

An easy computation gives $D:=T^{-1}AT=\left(\begin{matrix}-1 & \ \ 0 \\ \ \ 0 & -9 \\\end{matrix}\right)$ As $D$ is diagonal, (2) is naturally an uncoupled system: setting $Y=\binom{y_1(t)}{y_2(t)}$, it becomes

$$\tag{3}\ddot y_1=-y_1 \ \ and \ \ \ddot y_2=-9y_2(t).$$

As $Y=T^{-1}X=\dfrac12 TX$, the initial values are clearly

$y_1(0)=\tfrac12(x_1(0)+x_2(0))=1$ and $y_2(0)=\tfrac12(x_1(0)-x_2(0))=1$ and

$\dot y_1(0)=\tfrac12(\dot x_1(0)+\dot x_2(0))=0$ and $\dot y_2(0)=\tfrac12(\dot x_1(0)-\dot x_2(0))=-1.$

Now, because (3) gives $y_1(t)=A \cos(t)+B\sin(t)$ and $y_2(t)=C \cos(3t)+D\sin(3t)$, it is easy to find $A,B,C,D$ by taking the initial values just found upwards. I leave it to you...

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