What are the roots of the characteristic equation 2m^2 + (1-1)m - 1 = 0?

[ryan8888]

Homework Statement

Solve: 2x²y" + xy' - y = 3x⁴

Homework Equations

The Attempt at a Solution

I know this is a Euler Cauchy solution because of the x² before the y" in the ODE.

I try to solve the characteristic equation m² +(a-1)m + b = 0

I get 2m² + (1-1)m - 1 =0 or 2m² - 1 = 0

What I tried was:

2(m²-1) = 0
So one root would be m = 1

But I still don't think that is the correct solution

 

[LCKurtz]

Your characteristic equation is wrong. Instead of plugging into some formula, try substituting y = x^m into your DE and get the correct characteristic equation for yourself.

 

[ryan8888]

Okay thanks. I didn't realize that you couldn't just simply utilize that formula for the characteristic equation.

So I took y = x^m and found its derivatives:

y' = mx^m-1
y" = m(m-1)x^m-2

Substituting them into my original ODE and simplifying I obtain the following characteristic equation:

2m² - m - 1 = 0

This is from an old test I've been practicing on. Is that the correct characteristic equation?

Your characteristic equation is wrong. Instead of plugging into some formula, try substituting y = x^m into your DE and get the correct characteristic equation for yourself.

 

[LCKurtz]

Okay thanks. I didn't realize that you couldn't just simply utilize that formula for the characteristic equation.

So I took y = x^m and found its derivatives:

y' = mx^m-1
y" = m(m-1)x^m-2

Substituting them into my original ODE and simplifying I obtain the following characteristic equation:

2m² - m - 1 = 0

This is from an old test I've been practicing on. Is that the correct characteristic equation?

Correct. And much more reliable than trying to remember another formula.

 

[ryan8888]

Correct. And much more reliable than trying to remember another formula.

Okay 1 final question. When I go to determine my roots what is the best way to handle the 2m^2? Is it the same as saying two numbers that multiply to -2 and add to -1. So my roots are m = 2 and m = -1?

And I agree the y = x^m method is significantly easier to deal with!

Thanks

 

[LCKurtz]

Okay 1 final question. When I go to determine my roots what is the best way to handle the 2m^2? Is it the same as saying two numbers that multiply to -2 and add to -1. So my roots are m = 2 and m = -1?

And I agree the y = x^m method is significantly easier to deal with!

Thanks

Oh my...here you are in differential equations and are asking how to solve a quadratic equation? Factor it. Or failing that, use the quadratic formula.