# Let $y(t)=(y_1,y_2)^t$ satisfy $\frac{dy}{dt}=Ay,t>0$ such that..

I came across the above problem and my attempts are as follows: We see the characteristic polynomial of $A$ is $t^2-{\rm trace} (A)t+\det(A)=0$ and hence $t^2- \lambda^2=0$ since ${\rm trace}(A)=0$ and $-\lambda^2=\det(A)>0$. Suppose $\lambda^2=-\alpha^2<0$ and so $\lambda=\pm i\alpha$. Thus $$y(t)=C_1\cos(\alpha t)+C_2\sin(\alpha t)$$ Now I do not know how to use the fact $y(0)=(0,1)^t$ . Can someone point me in the right direction?

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You have $y(t)$ as a scalar function, when the $y(t)$ in your DE is a vector. –  JohnD Jan 30 '13 at 3:06
Can you explain more elaborately,sir? How can I use this fact to come to any conclusion? –  user53386 Jan 30 '13 at 3:34
You write $y(t)=C_1\cos(\alpha t)+C_2\sin(\alpha t)$ which is clearly a scalar function, not a vector. Yet the problem statement says $y(t)=(y_1(t),y_2(t))^T$ where I am using $^T$ for transpose. Also, the given initial condition is $y(0)=(0,1)^T$ which gains says that $y(t)$ is a vector-valued function. –  JohnD Jan 30 '13 at 3:38
So,using the fact $y(0)=(0,1)^t$ we get $y(0)=C_1=(0,1)^t$ and the value of $C_2$ remains undetermined. –  user53386 Jan 30 '13 at 3:51

Since the eigenvalues are pure imaginary, $\lambda=\pm\alpha i$, the general solution is of the form $$\mathbf{y}(t)=c_1\mathbf{v}_1\cos(\alpha t)+c_2\overline{\mathbf{v}}_1\sin(\alpha t).\tag{1}$$

This immediately rules out (1) and (2).

If (4) were true, then $c_1=c_2=0$, but then the initial condition $\mathbf{y}(0)=\begin{bmatrix}0\\1\end{bmatrix}$ is violated.

Hence, the answer is (3), and this makes sense based on the form of $(1)$.

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You've done a nice work. Even if it is not possible to determine explicitly the two vectors $\mathrm C_1$ and $\mathrm C_2$, this give you the indication that the answer lies in $\{3, 4\}$. The only thing to prove now is that those two vectors cannot be zero.

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