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I have this system

$$\begin{bmatrix}\frac{dy_1(x)}{dx} \\ \frac{dy_2(x)}{dx} \end{bmatrix}=\begin{bmatrix} \frac{1}{x} & \frac{1}{x} \\ \frac{4}{x} & \frac{1}{x} \end{bmatrix}\begin{bmatrix}y_1(x) \\ y_2(x) \end{bmatrix}$$

I can solve system of ODE with constant coefficients and write the general solution but this is not the case. Does this system have a solution you can write with elementary functions? If this is the case how can I do? Is there a systematic procedure? Could you explain me?

Any help will be appreciated.

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  • $\begingroup$ You have to make a change of variable. $\endgroup$ – Kel Jan 28 '18 at 1:02
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    $\begingroup$ How can I do please? $\endgroup$ – Helb Jan 28 '18 at 1:07
  • $\begingroup$ I'm writing an answer $\endgroup$ – Kel Jan 28 '18 at 1:08
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In general for a system that can be written in the form

$$\frac{d}{dx} \overline{y}(x)=f'(x)A\,\overline{y}(x)$$

where $\overline{y}$ is a vector, $A$ is a matrix of constant coefficients and $f'(x)$ is the derivative of a function $f$ you can proceed by making the change of variable

$$z=f(x)$$

with differential

$$dz=f'(x)dx$$

and

$$\frac{dx}{dz}=\frac{1}{f'(x)}$$ Using this change of variable you can write the initial system as a function of $z$ as

$$\frac{dx}{dz}\frac{d}{dx} \overline{y}(z)=\frac{dx}{dz}f'(x)A\,\overline{y}(z)$$

but simplifying the differentials in the left and using that $\frac{dx}{dz}=\frac{1}{f'(x)}$ in the right you get a system of ODEs with constant coefficients in the variable $z$

$$\frac{d}{dz} \overline{y}(z)=A\,\overline{y}(z)$$

that you can solve and you will find the solution $y_{\rm sol}(z)$. The solution in the initial variable $x$ will be found by using the change of variable $z=f(x)$ so that the final solution will be

$$y_{\rm sol}(f(x))$$

In your specific case you can rewrite the system like

$$\frac{d}{dx}\begin{bmatrix} y_1(x) \\ y_2(x) \end{bmatrix}=\frac{1}{x}\begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}\begin{bmatrix}y_1(x) \\ y_2(x) \end{bmatrix}$$ Then, you make a change of variable $$z=\ln x$$ hence the differential $$dz=\frac{1}{x}dx$$ and $$\frac{dx}{dz}=x$$ Using this change of variable you can write the initial system as a function of $z$ as $$\frac{dx}{dz}\frac{d}{dx}\begin{bmatrix} y_1(z) \\ y_2(z) \end{bmatrix}=\frac{dx}{dz}\frac{1}{x}\begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}\begin{bmatrix}y_1(z) \\ y_2(z) \end{bmatrix}$$ but simplifying the differentials in the left and using that $\frac{dx}{dz}=x$ in the right you get a system of ODEs with constant coefficients in the variable $z$

$$\frac{d}{dz}\begin{bmatrix} y_1(z) \\ y_2(z) \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}\begin{bmatrix}y_1(z) \\ y_2(z) \end{bmatrix}$$ that you can solve and you will find the solution $y_{\rm sol}(z)$. The solution in the initial variable $x$ will be found by using the change of variable $z=\ln x$ so that the final solution will be

$$y_{\rm sol}(\ln x)$$

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Let $A=\begin{bmatrix}1 & 1 \\ 4 & 1\end{bmatrix}$, $y=\begin{bmatrix}y_1 \\ y_2\end{bmatrix}$ and write the equation as $$ \frac{dy}{dx}-\frac{1}{x}Ay = 0. $$ The characteristic polynomial of $A$ is $(\lambda-1)^2-4=(\lambda-3)(\lambda+1)$. So $A$ diagonalizable. It has a basis of eigenvectors $$ X_1=\begin{bmatrix}1 \\ 2\end{bmatrix}, \; X_2 =\begin{bmatrix}-1 \\ 2\end{bmatrix} $$ $AX_1 = 3X_1$ and $AX_2 = -X_2$.Then $y(x)=a(x)X_1+b(x)X_2$ can be substituted into the original equation to obtain $$ a'(x)X_1+b'(x)X_2-\frac{3}{x}a(x)X_1+\frac{1}{x}b(x)X_2=0. $$ This gives component equations $$ a'(x)-\frac{3}{x}a(x)=0,\;\; b'(x)+\frac{1}{x}b(x)=0 \\ \frac{a'}{a}=\frac{3}{x},\;\; \frac{b'}{b}=-\frac{1}{x} \\ a = Ax^3,\;\; b=B\frac{1}{x}. $$ The general solution is $$ y(x) = Ax^3\begin{bmatrix}1 \\ 2\end{bmatrix}+B\frac{1}{x}\begin{bmatrix}-1 \\ 2\end{bmatrix}, $$ where $A$ and $B$ are arbitrary constants. So $y_1(x)=Ax^3-B/x$ and $y_2(x)=2Ax^3+2B/x$ are your solutions.

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