- #1

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**Summary::**We want to find explicit functions ##g(y,t)## and ##f(y,t)## satisfying the following system of differential equations.

I attached a very similar solved example.

Given the following system of differential equations (assuming ##y \neq 0##)

\begin{equation*}

-y\partial_t \left( g(y,t)\right) + f(y,t) = 0 \tag{1}

\end{equation*}

\begin{equation*}

y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2}

\end{equation*}

\begin{equation*}

\frac{1}{y^2} \partial_t \left( f(y,t) \right) - \partial_y \left( \frac{1}{y^2} g(y,t) \right) - \frac{2}{y^3} g(y,t) = 0 \tag{3}

\end{equation*}

We want to find explicit functions ##g(y,t)## and ##f(y,t)## satisfying such system

Adding up ##(1)## and ##(2)##

\begin{equation*}

\partial_y \left( f(y,t)\right) - \partial_t \left( g(y,t)\right) = 0 \tag{4}

\end{equation*}

Equation ##(4)## is satisfied when

\begin{equation*}

f(y,t) = f(t), \ \ \ \ g(y,t) = g(y) \tag{5}

\end{equation*}

Plugging ##(5)## into ##(3)## yields

\begin{equation*}

\partial_t f(t) - \partial_y g(y) =0 \tag{6}

\end{equation*}

But I do not see how ##(6)## is going to lead us to get explicit functions for ##g(y,t)## and ##f(y,t)##.

I actually expect to get a second order differential equation out of working out the system, whose solution should provide us with explicit functions for ##g(y,t)## and ##f(y,t)##.

Could you please help me out to move forward?

Thank you!

PS: I attach an analogous example. I am following the same procedure to solve our original problem