How to solve this system of differential equations? it seems that $x$ is depending from $t$?

Question

Solve the system:

$$ \begin{cases} tx'(t)=x(t)+y(t),\\ ty'(t)=3x(t)-y(t). \end{cases} $$

No initial conditions were given or anything like that, that's it.

Answer 1

we get $$tx''=x'+y'$$ so we get $$tx''=x'+\frac{3}{t}x-\frac{1}{t}y$$ and with your first equation $$tx''=x'+\frac{3}{t}x-\frac{1}{t}(tx'-x)$$

Answer 2

Adding the two equations yields $$t(x'+y')=4x$$ Differentiating the first equation yields $$x'+tx''=x'+y'$$ Using the second to substitute for $x'+y'$ in the first yields $$t(x'+tx'')=4x$$ which can be solved in the form $x=ax^{n_1}+bx^{n_2}$

Answer 3

Avoiding any tricks, the best is clearly to diagonalize the matrix on the right.

Note that $$ S^{-1}\begin{pmatrix}1 & 1\\ 3 & -1\end{pmatrix}S=\begin{pmatrix}-2 & 0\\ 0 & 2\end{pmatrix}, \quad\text{where}\quad S=\begin{pmatrix}-1 & 1\\ 3 & 1\end{pmatrix} $$ (simply compute the eigenvalues and eigenvectors). So in the new variables $$ \begin{pmatrix} z\\w\end{pmatrix}=S^{-1}\begin{pmatrix} x\\y\end{pmatrix}, \tag 1 $$ the original equation $$ t\begin{pmatrix} x\\y\end{pmatrix}'=\begin{pmatrix}1 & 1\\ 3 & -1\end{pmatrix}\begin{pmatrix} x\\y\end{pmatrix} $$ can be written in the form $$ t\begin{pmatrix} z\\w\end{pmatrix}'=\begin{pmatrix}-2 & 0\\ 0 & 2\end{pmatrix}\begin{pmatrix} z\\w\end{pmatrix}. $$ So you need to solve $$ tz'=-2z\quad\text{and}\quad tw'=2w. $$ After that you can recover $x$ and $y$ from $(1)$.

Answer 4

Introduce a new variable $\tau=\ln{t}$, then $$t\frac{d}{dt}=\frac{d}{d\tau}$$ Thus you get a linear system with constant coefficients in variable $\tau$. I assume you know how to proceed...