Chapter 4: Problem 6
Reduce the equation $$ x^{2}\left(\frac{d^{2} y}{d x^{2}}\right)+2 \pi\left(\frac{d y}{d x}\right)-5 y=0 $$ to a differential equation with constant coefficients in \(d^{2} y / d z^{2}, d y / d z_{1}\) and \(y\) by the change of variable \(x=e^{\mathrm{r}}\)
Short Answer
Expert verified
\[ \frac{d^2 y}{d r^2} + 2 \pi \frac{d y}{d r} - 5y = 0 \]
Step by step solution
01
Identify the change of variable
The given problem suggests the change of variable: \[ x = e^r \]. Therefore, express derivatives in terms of the new variable \( r \).
02
Transform the first derivative
Using the chain rule, \[ \frac{d y}{d x} = \frac{d y}{d r} \cdot \frac{d r}{d x} \]. Since \( r = \ln x \), we get \( \frac{d r}{d x} = \frac{1}{x} \). Hence, \[ \frac{d y}{d x} = \frac{1}{x} \frac{d y}{d r} \].
03
Transform the second derivative
Applying the chain rule again for the second derivative:\[ \frac{d^2 y}{d x^2} = \frac{d}{d x} \left( \frac{1}{x} \frac{d y}{d r} \right) = \frac{d}{d x} \left( \frac{1}{x} \right) \frac{d y}{d r} + \frac{1}{x} \frac{d}{d x} \left( \frac{d y}{d r} \right) \]. Since \( \frac{d}{d x} \left( \frac{d y}{d r} \right) = \frac{d^2 y}{d r^2} \cdot \frac{d r}{d x} = \frac{1}{x} \frac{d^2 y}{d r^2} \), we get:\[ \frac{d^2 y}{d x^2} = \left( -\frac{1}{x^2} \frac{d y}{d r} + \frac{1}{x^2} \frac{d^2 y}{d r^2} \right) \].
04
Substitute derivatives into the original equation
Replace the derivatives in the original equation with their corresponding expressions in \( r \): \[ x^2 \left( -\frac{1}{x^2} \frac{d y}{d r} + \frac{1}{x^2} \frac{d^2 y}{d r^2} \right) + 2\pi \left( \frac{1}{x} \frac{d y}{d r} \right) - 5y = 0 \].
05
Simplify the equation
Multiply through by \( x^2 \) to eliminate denominators: \[ - \frac{d y}{d r} + \frac{d^2 y}{d r^2} + 2 \pi x \frac{d y}{d r} - 5x^2 y = 0 \].
06
Correct the form with respect to \( x = e^r \)
Since \( x = e^r \), we simplify terms accordingly. We get:\[ \frac{d^2 y}{d r^2} + 2 \pi \frac{d y}{d r} - 5y = 0 \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Change of Variables
A change of variables is an essential technique for simplifying differential equations. In this exercise, we start with the substitution \(x = e^r\). This substitution is helpful because it transforms our variable into a more manageable form, making the equation easier to work with.
By changing variables, we often turn a complicated problem into one that is simpler or has constant coefficients, as we'll see in this solution.
This approach can be powerful, especially when the original variable makes the equation complicated.
First, identify the new variable and express it mathematically. Here we used \(x = e^r\).
Next, express all derivatives in terms of this new variable. In this case, we need both the first and second derivatives.
By changing variables, we often turn a complicated problem into one that is simpler or has constant coefficients, as we'll see in this solution.
This approach can be powerful, especially when the original variable makes the equation complicated.
First, identify the new variable and express it mathematically. Here we used \(x = e^r\).
Next, express all derivatives in terms of this new variable. In this case, we need both the first and second derivatives.
Constant Coefficients
One goal of this exercise is to reduce the original equation to one with constant coefficients. Equations with constant coefficients are simpler to solve because they don't change with the independent variable.
After performing the change of variables, we need to substitute these new expressions back into our original equation.
For instance, the initial term involving \(x^2\) can become simpler by adopting a new form of the derivatives.
When substituting back, make sure to multiply out terms to eliminate denominators, creating a clearer equation.
In our final form, \(\frac{d^2 y}{d r^2} + 2\pi \frac{d y}{d r} - 5y = 0\), each coefficient in terms of \(y\) and its derivatives is constant.
This makes the differential equation easier to solve or analyze further.
After performing the change of variables, we need to substitute these new expressions back into our original equation.
For instance, the initial term involving \(x^2\) can become simpler by adopting a new form of the derivatives.
When substituting back, make sure to multiply out terms to eliminate denominators, creating a clearer equation.
In our final form, \(\frac{d^2 y}{d r^2} + 2\pi \frac{d y}{d r} - 5y = 0\), each coefficient in terms of \(y\) and its derivatives is constant.
This makes the differential equation easier to solve or analyze further.
Chain Rule
The chain rule is the backbone of transforming variables and derivatives. It connects how a function changes with respect to one variable to how it changes with respect to another variable.
In this example, we started with \(\frac{d y}{d x}\) and used the chain rule to express it in terms of \(r\): \(\frac{d y}{d x} = \frac{d y}{d r} \cdot \frac{d r}{d x}\).
Since \(r = \ln x\), we find \(\frac{d r}{d x}\), which equals \(\frac{1}{x}\).
The first derivative becomes \(\frac{d y}{d x} = \frac{1}{x} \frac{d y}{d r}\).
The second derivative requires applying the chain rule again: \(\frac{d^2 y}{d x^2}\).
This derivative involves both the first and second derivatives with respect to \(r\) and simplifies to involve terms like \(\frac{d y}{d r}\) and \(\frac{d^2 y}{d r^2}\).
Understanding these transformations and applications of the chain rule is key to mastering change of variables in differential equations.
In this example, we started with \(\frac{d y}{d x}\) and used the chain rule to express it in terms of \(r\): \(\frac{d y}{d x} = \frac{d y}{d r} \cdot \frac{d r}{d x}\).
Since \(r = \ln x\), we find \(\frac{d r}{d x}\), which equals \(\frac{1}{x}\).
The first derivative becomes \(\frac{d y}{d x} = \frac{1}{x} \frac{d y}{d r}\).
The second derivative requires applying the chain rule again: \(\frac{d^2 y}{d x^2}\).
This derivative involves both the first and second derivatives with respect to \(r\) and simplifies to involve terms like \(\frac{d y}{d r}\) and \(\frac{d^2 y}{d r^2}\).
Understanding these transformations and applications of the chain rule is key to mastering change of variables in differential equations.