Trouble with a second order differential equation

[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=e^z to find the general solution.

b,c are real, and b²>c

Then, using the special solution y=x^p, 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*x^r₁ + c*x^r₂

So then solving the special solution

x²*(p-1)*(p)*x^p-2+x*p*x^p-1+c*x^p = 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?

 

[JThompson]

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

 

[draco193]

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

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

 

[JThompson]

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

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

?

 

[draco193]

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.