# Solving a system of differential equations

• JD_PM
In summary, the conversation discusses finding explicit functions ##g(y,t)## and ##f(y,t)## that satisfy a given system of differential equations. The equations involve partial derivatives and the goal is to find solutions for ##g(y,t)## and ##f(y,t)## that satisfy the system. The conversation includes a solved example and discusses various approaches to solve the system. Ultimately, the goal is to find explicit functions for ##g(y,t)## and ##f(y,t)## that satisfy the system.

#### JD_PM

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

\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)##.

Thank you!

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

Delta2
Is this homework ?

JD_PM said:
Equation (4) is satisfied when
But then (2) yields ##f = 0## !

By the way, (2) is easily solved for ##y## !

BvU said:
Is this homework ?

This is a system of differential equations I encountered while solving a specific problem I am solving to study for my exam. I guess you can call it homework

BvU said:
But then (2) yields ##f = 0## !

By the way, (2) is easily solved for ##y## !

So if I am not mistaken you are suggesting not to add up ##(1)## and ##(2)## (?) How to proceed then? (sorry I am a bit rusty with differential equations).

JD_PM said:
I guess you can call it homework

JD_PM
JD_PM said:
sorry I am a bit rusty with differential equations
JD_PM said:
solving a specific problem I am solving to study for my exam
If (2) is already a stumbling block, and the 'analogous example' is typical for the exam, I fear with great fear ...

Compare
$$\begin{equation*} y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2} \end{equation*}$$
with ##\ x \, f'(x) = f(x) \ ##

JD_PM
BvU said:
If (2) is already a stumbling block, and the 'analogous example' is typical for the exam, I fear with great fear ...

... let's not panic though!

BvU said:
Compare
$$\begin{equation*} y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2} \end{equation*}$$
with ##\ x \, f'(x) = f(x) \ ##

Thanks for the guidance! Let's start over

Focusing on ##(2)##, we deduce that it takes the form

\begin{equation*}
\partial_y \left( y \ f(x,y) \right) = 0
\end{equation*}

Where we need to impose

\begin{equation*}
\partial_y f(x,y) = - f(x,y)
\end{equation*}

The simplest function I can think of that satisfies this equation (setting the constant of integration equal to zero) is ##f(x,y) = e^{-\ln(y)}## i.e.

\begin{equation*}
\partial_y \left( y \ e^{-\ln(y)} \right) = e^{-\ln(y)} - y\left( \frac{1}{y} \right) \ e^{-\ln(y)} = \frac{1}{y} - \frac{1}{y} = 0
\end{equation*}

Plugging ##f(x,y) = e^{-\ln(y)}## into ##(1)## yields

\begin{equation*}
-y\partial_t \left( g(y,t) \right) + e^{-\ln(y)} = 0 \Rightarrow \partial_t \left( g(y,t) \right) = \frac{1}{y^2}
\end{equation*}

Thus (setting the constant of integration equal to zero) we get that ##g(x,y) = t/y^2##

But I have missed something as ##(3)## is not satisfied due to a factor of ##2## i.e.

\begin{equation*}
-\partial_y \left( \frac{t}{y^4} \right) - \frac{2t}{y^5} = \frac{2t}{y^5} \neq 0
\end{equation*}

What am I missing? I guess that I should impose a specific initial condition to get it done but I do not see it...

##\qquad##Funny, from $$\begin{equation*} \partial_y \left( y \; f(x,y) \right) = 0 \end{equation*}$$##\qquad## I get $$\begin{equation*} y\,\partial_y \left( f(y,t)\right) +f(y,t) = 0 \end{equation*}$$ where ##(2)## has a minus sign !

By the way, we physicists usually write ##e^{-\ln y} \ ## as ##\ \displaystyle{1\over y}\ ##

JD_PM said:
... let's not panic though!
That's the spirit !

JD_PM
OK let's make it simpler

BvU said:
Compare
$$\begin{equation*} y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2} \end{equation*}$$
with ##\ x \, f'(x) = f(x) \ ##

So we see that ##f(y,t) = f(y) = e^{\ln y} = \frac{1}{y}## (so everyone is happy! haha) is a solution for ##(2)##

Then all we need to do is plugging it into ##(1)## and find ##g(y,t)## so that the following equation is satisfied

\begin{equation*}
\partial_t \left( g(y,t)\right) = \frac{1}{y^2}
\end{equation*}

Besides, ##g(y,t)## and ##f(y,t)## must be a solution for ##(3)## of course.

Thank you, I think I can get it done now (time to go back to study! ).

JD_PM said:
so everyone is happy! haha
Hoho! Not yet! I'll let you know when I'm happy

As long as you write
JD_PM said:
So we see that ##f(y,t) = f(y) = e^{\ln y} = \frac{1}{y}##
I'm not happy by a long shot ! And the last bit ##e^{\ln y} = \frac{1}{y} ## is (probably ?) a typo, so that does not count.

##f(y,t)=f(y) ## is the one that is really irking me. We solve (we try to ...) for ##y## but that doesn't mean ##t## goes away.

We are still working on ##(2)## and are nearly there ...

JD_PM
BvU said:
Hoho! Not yet! I'll let you know when I'm happy

As long as you write
I'm not happy by a long shot ! And the last bit ##e^{\ln y} = \frac{1}{y} ## is (probably ?) a typo, so that does not count.

Oops a typo indeed. Let's go slowly

I meant that a solution for ##(2)## is given by

\begin{equation*}
f(y,t) = e^{\ln y} = y
\end{equation*}

Next step would be going to ##(1)## and solve for ##g(y,t)##, wouldn't it?

No: any function of ##t## can be factored in in both terms of ##(2)## without affecting the validity of the equation.

BvU said:
By the way, we physicists usually write ##e^{-\ln y} \ ## as ##\ \displaystyle{1\over y}\ ##
As do mathematicians...

BvU said:
No: any function of ##t## can be factored in in both terms of ##(2)## without affecting the validity of the equation.

I better take more time to study and then reply. I do not want to exhaust your patience

## 1. What is a system of differential equations?

A system of differential equations is a set of equations that describe the relationship between a set of variables and their rates of change. These equations are often used in mathematical models to predict how a system will change over time.

## 2. How do you solve a system of differential equations?

There are several methods for solving a system of differential equations, including analytical methods and numerical methods. Analytical methods involve finding an exact solution using algebraic techniques, while numerical methods involve using algorithms to approximate the solution.

## 3. What is the importance of solving a system of differential equations?

Solving a system of differential equations allows us to make predictions about the behavior of complex systems, such as weather patterns or population dynamics. It also allows us to understand the relationships between different variables in a system and how they change over time.

## 4. What are some common techniques used to solve a system of differential equations?

Some common techniques include separation of variables, substitution, and the use of integrating factors. Other methods, such as Laplace transforms and power series, can also be used for more complex systems.

## 5. Are there any real-world applications of solving systems of differential equations?

Yes, there are many real-world applications, such as predicting the spread of diseases, modeling chemical reactions, and understanding the behavior of electrical circuits. These equations are also used in engineering, physics, and economics to make predictions and solve problems.