- #1

draco193

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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.

I am given the second order differential equation

x

Use the transform x=e

b,c are real, and b

Then, using the special solution y=x

The find the real solutions if b

The Quadratic Formula will be used

For the first part, transforming the original equation:

y'(x) = [tex]\frac{dy}{dz}[/tex]*[tex]\frac{dz}{dx}[/tex] =y'(z)*[tex]\frac{1}{x}[/tex]

similarly, y''(x) = [tex]\frac{-1}{x^2}[/tex].*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

[tex]\frac{(2b+1) \mp sqrt((4b^2)+4b+4c+1}{2}[/tex]

Using these as r

So then solving the special solution

x

Simplifying, I get

p

Solving for p, I get p = -b[tex]\pm[/tex] sqrt(b

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

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

x

^{2}*y''(x)+(2*b+1)*x*y'(x)+c*y(x) = 0Use the transform x=e

^{z}to find the general solution.b,c are real, and b

^{2}>cThen, using the special solution y=x

^{p}, solve for p and confirm you get the same solution.The find the real solutions if b

^{2}=c, b^{2}<c## Homework Equations

The Quadratic Formula will be used

## The Attempt at a Solution

For the first part, transforming the original equation:

y'(x) = [tex]\frac{dy}{dz}[/tex]*[tex]\frac{dz}{dx}[/tex] =y'(z)*[tex]\frac{1}{x}[/tex]

similarly, y''(x) = [tex]\frac{-1}{x^2}[/tex].*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

[tex]\frac{(2b+1) \mp sqrt((4b^2)+4b+4c+1}{2}[/tex]

Using these as r

_{1}and r_{2}, and replacing z with x, I find the general solution y(x) = c*x^{r1}+ c*x^{r2}So then solving the special solution

x

^{2}*(p-1)*(p)*x^{p-2}+x*p*x^{p-1}+c*x^{p}= 0Simplifying, I get

p

^{2}+2bp+c =0Solving for p, I get p = -b[tex]\pm[/tex] sqrt(b

^{2}-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?

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