Characteristic equation with x^2 coefficient

[knockout_artist]

Homework Statement

x² d²y/dx² + 3x dy/dx + 5y = g(x)

Homework Equations

How do we find Characteristic equation for it.

The Attempt at a Solution

x²λ² + 3xλ + 5 = 0
λ₁ = 1/2 [-x² + √ (x⁴ + 20 ) ]
λ₂ = 1/2[ -x² - √(x⁴ + 20) ]

I used 1/3 -/+ a √(a² + 4b)
where
a = x²
b = 5

 

[Mark44]

Homework Statement

x² d²y/dx² + 3x dy/dx + 5y = g(x)

Homework Equations

How do we find Characteristic equation for it.

If memory serves, the characteristic equation applies to the related nonhomogeneous DE -- $x^2 y'' + 3x y' + 5y = 0$.
This is an example of an 2nd order Euler equation. One technique is to assume a solution of the form $y = x^n$, and substitute it into the DE. For your nonhomogeneous equation, it depends on what g(x) is.
See https://www.math24.net/second-order-euler-equation/

The Attempt at a Solution

x²λ² + 3xλ + 5 = 0
λ₁ = 1/2 [-x² + √ (x⁴ + 20 ) ]

I have no idea what you're doing here.

λ₂ = 1/2[ -x² - √(x⁴ + 20) ]

I used 1/3 -/+ a √(a² + 4b)
where
a = x²
b = 5

 

[Ray Vickson]

Homework Statement

x² d²y/dx² + 3x dy/dx + 5y = g(x)

Homework Equations

How do we find Characteristic equation for it.

The Attempt at a Solution

x²λ² + 3xλ + 5 = 0
λ₁ = 1/2 [-x² + √ (x⁴ + 20 ) ]
λ₂ = 1/2[ -x² - √(x⁴ + 20) ]

I used 1/3 -/+ a √(a² + 4b)
where
a = x²
b = 5

Forget characteristic equations: they do not apply in this problem. They are for constant coefficient linear DEs, but your DE has coefficients that are functions of $x$.

 

[Mark44]

Forget characteristic equations: they do not apply in this problem.

The old saying applies here: "If the only tool you have is a hammer, everything looks like a nail."
As Ray said (and I forgot), characteristic equations are applicable only to constant coefficient linear differential equations, and specifically to homogeneous DEs of that type.

 

[MidgetDwarf]

If you need a quick reference, the method is called cauchy-euler (I think this is the name). There is also an equivalent form using a log, worth it to know both versions. t I believe you can also use Frobineous Method, due to the analytical point. Its been years since I solved a differential equation.