# Trouble with a second order differential equation

• draco193
In summary, the conversation discusses solving a second order differential equation using the transform x=ez and finding the general solution, as well as solving for a special solution and finding the real solutions under certain conditions. A mistake is identified in the calculation of y''(x) and is corrected.
draco193
Hello all,

I'm having a little trouble getting the Latex to work, so I'm hoping this won't be too hard for everyone to understand.

## Homework Statement

I am given the second order differential equation
x2*y''(x)+(2*b+1)*x*y'(x)+c*y(x) = 0

Use the transform x=ez to find the general solution.

b,c are real, and b2>c

Then, using the special solution y=xp, solve for p and confirm you get the same solution.

The find the real solutions if b2=c, b2<c

## Homework Equations

The Quadratic Formula will be used

## The Attempt at a Solution

For the first part, transforming the original equation:

y'(x) = $$\frac{dy}{dz}$$*$$\frac{dz}{dx}$$ =y'(z)*$$\frac{1}{x}$$

similarly, y''(x) = $$\frac{-1}{x^2}$$.*y''(z)

Replacing into the original equation, I get
-y''(z)+(2b+1)*y'(z)+c*y(z).

Using the quadrtaic equation, I then get as a solution
$$\frac{(2b+1) \mp sqrt((4b^2)+4b+4c+1}{2}$$

Using these as r1 and r2, and replacing z with x, I find the general solution y(x) = c*xr1 + c*xr2

So then solving the special solution

x2*(p-1)*(p)*xp-2+x*p*xp-1+c*xp = 0

Simplifying, I get

p2+2bp+c =0

Solving for p, I get p = -b$$\pm$$ sqrt(b2-c)

This obviously does not equal my other solution, so I'm wondering if anyone can see where it is I've made my mistake?

Last edited:
You need to calculate y''(x) not y''(z).

JThompson said:
You need to calculate y''(x) not y''(z).

Thats what I get for trying to type my work in quickly. Typo fixed.

I got something different for y''(x). Did you calculate

$$\frac{d^{2}y}{d^{2}x}=\frac{d(\frac{dy}{dx})}{dz}*\frac{dz}{dx}$$

using the product rule for derivatives on

$$\frac{d(\frac{dy}{dx})}{dz}$$

?

JThompson said:
I got something different for y''(x). Did you calculate

$$\frac{d^{2}y}{d^{2}x}=\frac{d(\frac{dy}{dx})}{dz}*\frac{dz}{dx}$$

using the product rule for derivatives on

$$\frac{d(\frac{dy}{dx})}{dz}$$

?

Thank you. That is where my error was at.

## 1. What is a second order differential equation?

A second order differential equation is a mathematical equation that involves the second derivative of a function. It is used to describe the relationship between a function and its rate of change, and is commonly used in physics and engineering.

## 2. What makes solving second order differential equations difficult?

The main difficulty in solving second order differential equations is that they often involve complex mathematical operations and require advanced calculus techniques. Additionally, the initial conditions and boundary conditions for the equation must be known in order to find a solution.

## 3. How is a second order differential equation different from a first order differential equation?

A second order differential equation involves the second derivative of a function, while a first order differential equation only involves the first derivative. This means that a second order differential equation is more complex and requires more information to solve.

## 4. What are some real-world applications of second order differential equations?

Second order differential equations are used in many fields, including physics, engineering, and economics. They can be used to model the motion of objects, the growth of populations, and the behavior of electrical circuits, among other things.

## 5. What are some techniques for solving second order differential equations?

There are several techniques for solving second order differential equations, including separation of variables, variation of parameters, and using power series solutions. Depending on the specific equation, different techniques may be more efficient or useful.

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