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I have given the system:

$$\left(\begin{array}{c} {y_1}'\left(t\right)\\ {y_2}'\left(t\right) \end{array}\right) = \left(\begin{array}{cc} -\frac{2}{t} & \frac{1}{t}\\ \frac{3}{t} & 0 \end{array}\right) \left(\begin{array}{c} {y_1}\left(t\right)\\ {y_2}\left(t\right) \end{array}\right)$$

Here I thought about the fundamental solution as: $$\left(\begin{array}{c} {y_1}\left(t\right)\\ {y_2}\left(t\right) \end{array}\right) = \left(\begin{array}{c} v_1&v_2\end{array}\right) \,\left(\begin{array}{c} \exp(\lambda_1\,t)\\ \exp(\lambda_2\,t) \end{array}\right)$$

Now with eigenvectors: $v_1 = \left(\begin{array}{c} 1\\-1\end{array}\right), v_2 = \left(\begin{array}{c} 1\\3\end{array}\right)$ and eigenvalues $\lambda_1 = \frac{-3}{t}, \lambda_2 = \frac{1}{t}$ I get:

$$\left(\begin{array}{c} {y_1}\left(t\right)\\ {y_2}\left(t\right) \end{array}\right) = \left(\begin{array}{c} 1 &1 \\-1&3\end{array}\right) \,\left(\begin{array}{c} \exp(-3)\\ \exp(1) \end{array}\right)$$

However the solution proposes: $$\left(\begin{array}{c} {y_1}\left(t\right)\\ {y_2}\left(t\right) \end{array}\right) = \left(\begin{array}{c} \dfrac{1}{t^3} &t \\-\dfrac{1}{t^3}&3\,t\end{array}\right)$$

I acknowledge there is a narrow connection, but I don't see right why...

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    $\begingroup$ $e^{\lambda t}$ only works for constant coefficients. $\endgroup$
    – user58697
    Jun 11, 2021 at 17:27

2 Answers 2

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Let $A = \begin{pmatrix} -2 & 1 \\ 3 & 0 \end{pmatrix} $ which has eigenvalues $\lambda_1 = -3$, $\lambda_2 = 1$ and eigenvectors $v_1 = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}, v_2 = \dfrac{1}{\sqrt{10}}\begin{pmatrix}1 \\ 3\end{pmatrix}$. So $A = U \;\mathrm{diag}(\lambda_1, \lambda_2) \;U^{-1}$, where $U = \begin{pmatrix} v_1 & v_2 \end{pmatrix}$. The given ODE is $y' = \dfrac{1}{t}A y$, which has the solution $$ y(t) = e^{A\int_1^t \frac{1}{s}ds} y(1)= t^A y(1) = U \;\mathrm{diag}(t^{\lambda_1}, t^{\lambda_2}) \;U^{-1}\;y(1).$$

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Hint: You can condense this system into a $2$nd order Cauchy-Euler equation:

$${y_2}'=\frac3t y_1 \implies {y_2}'' = -\frac3{t^2}y_1+\frac3t{y_1}'$$

$$\implies {y_2}''=-\frac3{t}{y_2}'+\frac3{t^2}y_2$$

$$\implies t^2{y_2}''+3t{y_2}'-3y_2=0$$

Solve this for $y_2$, then differentiate and multiply by $\frac t3$ to recover $y_1$.

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