Solving system of differential equations : Wolfram Alpha vs theorem

eigenvalues-eigenvectorsordinary differential equationssystems of equations

I am burning my brain finding the most correct way to solve a system of differential equations. Here is an example :

$$\begin{cases}
x'=5x-2y\\
y'=-x+6y
\end{cases}
$$

Let's $Y(t)=\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$.
I use a matrix $A=\begin{pmatrix} 5 & -2 \\ -1 & 6 \end{pmatrix}$, calculate its eigen values $\lambda_1=7,\lambda_2=4 $ and eigen vectors $ v_1=\begin{pmatrix} -1 \\ 1 \end{pmatrix}, v_2=\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

Now 2 options:

  • I can use a theorem and find the solution is $Y(t)=c_1 e^{\lambda_1t} u_1 + c_2 e^{\lambda_2t} u_2$

  • I can continue using linear algebra, calculate $P$, $D$ and $P^{-1}$ such that $A=P D P^{-1}$ where D is a diagonal matrix.
    $Y'=AY$ so $Y=Ce^{tA}=C P \begin{pmatrix} e^{7t} & 0\\ 0&e^{4t} \end{pmatrix} P^{-1}$

I don't find same results but both are OK (the second is used by Wolfram). Did I misunderstand something? Which one is the best option?

Thanks for you answer

Best Answer

If you were to write that last formula correctly as $$ Y(t)=e^{tA}C= P \begin{pmatrix} e^{7t} & 0\\ 0&e^{4t} \end{pmatrix} P^{-1}C,~~~C=Y(0), $$ then you get the correspondence with the first formula with $$ P^{-1}C=\pmatrix{c_1\\c_2}, $$ and the columns of $P$ being the eigenvectors of $A$.

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