# Fundamental Solution of system of Ode: $\dfrac{1}{x}$

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...

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

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 $$y(t) = e^{A\int_1^t \frac{1}{s}ds} y(1)= t^A y(1) = U \;\mathrm{diag}(t^{\lambda_1}, t^{\lambda_2}) \;U^{-1}\;y(1).$$
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 $$y_1$$.