Wronskian for {e^{3x}, e^{-x}, 2}

ODEs: Show that the set of functions {e^{3x}, e^{-x}, 2} is a linearly independent set. These functions are in the solution space of y''' -2y'' - 3y' = 0. We show linear independence by computing the Wronskian of the set. We also show linear independence by solving a system of linear equations.

Category:

Education

Tags:

• Mathematics
• ordinary
• differential
• equation
• Wronskian
• linear
• independence
• span
• determinant
• solution
• space
• dependence
• cofactor

[chromosome24]

So finding the Wronskian for a set of equations is similar to determining if a set of vectors are describable by an orthogonal basis. For this though you use orthogonality of functions instead of vectors. another question is the nth derivative of some function f(x) linearly dependent to f(x)? great, concise video btw.

[MathDoctorBob]

@chromosome24 Thanks! The Wronskian follows our procedure to determine if a set of vectors form a basis. For an orthogonal basis, we would need an inner product to test basis vectors. If we had an inner product on our solution space and could find a set of orthogonal functions, this would show linear independence without the Wronksian.

If the nth derivative of f(x) is LD to f(x), then f^n(x) + cf(x) = 0. So we can try f(x) = e^{rx} and solve for r. Also try other roots of r^n - c.

[LeavingCertMaths]

Thanks.

[MathDoctorBob]

@LeavingCertMaths You're welcome!  - Bob

[QuickSaintPat]

I picture this dude as a karate instructor with the way he holds that wooden pole

[MathDoctorBob]

@QuickSaintPat It's an artifact from my stick and knife days. I'm all about jiu-jitsu now. - Bob

[MathDoctorBob]

@outcastoredwall Sweep the leg!

[outcastoredwall]

This feel like math day at the Cobra Kai dojo.

[MrKafziel]

@SteDeRaver Nice reference Futurama for the win

[SteDeRaver]

Sweet Clyde, use variation of parameters and expand the Wronskian!

[MathDoctorBob]

@stevenmcconnon You're welcome! - Bob

[stevenmcconnon]

Thanks!