# Solving an Euler differential equation

• member 587159
In summary, the conversation involves solving a differential equation and verifying a solution. The equation is converted into a linear equation with constant coefficients and variation of constants is used to find the particular solution. However, there is a discrepancy in the last term of the equation, making it difficult to find a solution.
member 587159

## Homework Statement

Solve the differential equation ##(2x+1)^2y'' + (4x+2)y' - 4y = x^2##
Can someone verify whether my solution is correct?

## The Attempt at a Solution

We perform the substitution ##t = \ln|2x+1|##. Then, ##e^t = |2x+1|## and ##x = \pm(e^t -1)/2##

Without loss of generality, let's solve the equation for ##x = (e^t-1)/2##. (the other case is analogue)

We find ##\frac{dy}{dx} = 2e^{-t}\frac{dy}{dt}##

##\frac{d^2y}{dx^2} = -4e^{-2t}\frac{dy}{dt}+4e^{-2t}\frac{d^2y}{dt^2}##

Hence, the differential equation becomes:

##16y'' - 16y = (e^t-1)^2##

This is a linear differential equation with constant coefficients:

The homogeneous is given by:

##y_h = c_1e^t + c_2e^{-t}##

Let's apply variation of the constants to find ##y_p##, the particular solution.

##y_p = c_1(t)e^t + c_2(t)e^{-t}##

Then, we can find the functions ##c_1'## and ##c_2'## as solutions of the system:

##\begin{cases}c_1'(t)e^t + c_2'(t)e^{-t} = 0\\c_1'(t)e^t - c_2'(t)e^{-t} = (e^t-1)^2\end{cases}##

##\iff c_1'(t) = -c_2'(t) = \frac{(e^t-1)^2}{2e^t}##

Hence:

##c_1(t) = e^t/2 - t - e^{-t}/2 = -c_2(t)## (if we set the constant of integration to ##0##)

Hence, the solution is given by:

##y = y_p + y_h = c_1e^t + c_2e^{-t} -1 + 1/2e^{2t}-te^t + 1/2e^{-2t}+te^{-t}##

And by substituting back:

##y(x) = c_1(2x+1) + c_2\frac{1}{2x+1} + \frac{(2x+1)^2}{2} - \ln(2x+1)(2x+1) - 1 + \frac{1}{2(2x+1)^2} + \frac{\ln(2x+1)}{2x+1}##

Math_QED said:

## Homework Statement

Solve the differential equation ##(2x+1)^2y'' + (4x+2)y' - 4y = x^2##
Can someone verify whether my solution is correct?

## The Attempt at a Solution

We perform the substitution ##t = \ln|2x+1|##. Then, ##e^t = |2x+1|## and ##x = \pm(e^t -1)/2##

Without loss of generality, let's solve the equation for ##x = (e^t-1)/2##. (the other case is analogue)

We find ##\frac{dy}{dx} = 2e^{-t}\frac{dy}{dt}##

##\frac{d^2y}{dx^2} = -4e^{-2t}\frac{dy}{dt}+4e^{-2t}\frac{d^2y}{dt^2}##

Hence, the differential equation becomes:

##16y'' - 16y = (e^t-1)^2##

This is a linear differential equation with constant coefficients:

The homogeneous is given by:

##y_h = c_1e^t + c_2e^{-t}##

Let's apply variation of the constants to find ##y_p##, the particular solution.

##y_p = c_1(t)e^t + c_2(t)e^{-t}##

Then, we can find the functions ##c_1'## and ##c_2'## as solutions of the system:

##\begin{cases}c_1'(t)e^t + c_2'(t)e^{-t} = 0\\c_1'(t)e^t - c_2'(t)e^{-t} = (e^t-1)^2\end{cases}##

##\iff c_1'(t) = -c_2'(t) = \frac{(e^t-1)^2}{2e^t}##

Hence:

##c_1(t) = e^t/2 - t - e^{-t}/2 = -c_2(t)## (if we set the constant of integration to ##0##)

Hence, the solution is given by:

##y = y_p + y_h = c_1e^t + c_2e^{-t} -1 + 1/2e^{2t}-te^t + 1/2e^{-2t}+te^{-t}##

And by substituting back:

##y(x) = c_1(2x+1) + c_2\frac{1}{2x+1} + \frac{(2x+1)^2}{2} - \ln(2x+1)(2x+1) - 1 + \frac{1}{2(2x+1)^2} + \frac{\ln(2x+1)}{2x+1}##
What is preventing you from putting your formula for ##y(x)## into the DE to test if it works?

Math_QED said:
##\begin{cases}c_1'(t)e^t + c_2'(t)e^{-t} = 0\\c_1'(t)e^t - c_2'(t)e^{-t} = (e^t-1)^2\end{cases}##

##\iff c_1'(t) = -c_2'(t) = \frac{(e^t-1)^2}{2e^t}##
This doesn't look right.

member 587159
vela said:
This doesn't look right.

Thanks for pointing out!

What a pity that last LHS term is –4y instead of 4y, in which case the LHS would have been an exact second derivative. Sure you copied it out right.?

member 587159
epenguin said:
What a pity that last LHS term is –4y instead of 4y, in which case the LHS would have been an exact second derivative. Sure you copied it out right.?

I just checked and I made no typo, unfortunately.

Pity they didn't ask good question. I can't see any way to hammer that one into something so nice.

## 1. What is an Euler differential equation?

An Euler differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is named after the Swiss mathematician Leonhard Euler, who made significant contributions to the field of calculus.

## 2. How is an Euler differential equation solved?

An Euler differential equation can be solved using various methods, such as separation of variables, substitution, or integrating factors. The specific method used will depend on the structure and complexity of the equation.

## 3. What is the significance of solving an Euler differential equation?

Solving an Euler differential equation allows us to find the solution to a wide range of real-world problems in fields such as physics, engineering, and economics. It also helps us understand the behavior of systems and make predictions about their future states.

## 4. What are some common applications of Euler differential equations?

Euler differential equations are commonly used in physics to describe the motion of objects, in chemistry to model chemical reactions, and in economics to analyze market dynamics. They are also used in many other fields, including biology, engineering, and finance.

## 5. Are there any limitations or challenges in solving an Euler differential equation?

Yes, there are several limitations and challenges when it comes to solving Euler differential equations. These include the complexity of the equations, the need for initial conditions, and the possibility of encountering non-analytic solutions. Additionally, some equations may not have closed-form solutions, requiring numerical methods for approximation.

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