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.


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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