Help with variation of constants

[sunrah]

Homework Statement

solve the following differential equation:
t⁴x'' - 4t³t' + 6t²x = - 12t - 20

Homework Equations

substitution x(t) = tⁿ

The Attempt at a Solution

this is a Euler equation with the following general solution: x(t) = c₁t² + c₂t³ worked out using the above substitution.

The particular solution should be obtainable through variation of constants but I just get a nonsense result:

The wronksian = W = 3c₁c₂t⁴ - 2c₁c₂t⁴ = c₁c₂t⁴

therefore:

x(t) = - x_{1} \int \frac{x_{2}b(t)}{W} dt + x_{2} \int \frac{x_{1}b(t)}{W}dt = x_{1} \int \frac{c_{2}t^{3}(12t + 20)}{c_{1}c_{2}t^{4}}dt - x_{2} \int \frac{c_{1}t^{2}(12t + 20)}{c_{1}c_{2}t^{4}} dt
x(t) = \frac{x_{1}}{c_{1}} \int (12 + \frac{20}{t})dt - \frac{x_{2}}{c_{2}} \int (\frac{12}{t} + \frac{20}{t^{2}}) dt

the integration is trivial but definitely isn't a particular solution!

 

[HallsofIvy]

Using "variation of parameters" (I would not call it "variation of constants") we do NOT include the constants- they become the "parameters". Knowing that t^2 and t^3 are solutions to the associated homogeneous equation, we look for solutions for the entire equation of the form x(t)= t^2u(t)+ t^3v(t) where we have replaced the constants with the "parameters" u and v. There are many such solutions- given any solution, we could find u and v to work.

Differentiating, we have x'= 2tu+ t^2u'+ 3t^2v+ t^3v'. Because there are many such solutions we "narrow the search" and simplify the calculations, by requiring that t^2u'+ t^3v'= 0. That leaves x'= 2tu+ 3t^2v. Differentiating again, x''= 2u+ 2tu'+ 6tv+ 3t^2v'.

Putting those into the original equation,
t^4x''- 4t^3x'+ 6t^2x= 2t^4u+ 2t^5u'+ 6t^5v+ 3t^6v'- 8t^4u- 12t^5v+ 6t^4u+ 6t^5v= -12t- 20
2t^4u- 9t^4u+ 6t^4u= 0 and 6t^5v- 12t^5v+ 5t^6v= 0 so the equation reduces to
2t^5u'+ 3t^6v'= -12t- 20
That, together with t^2u'+ t^3v'= 0 gives us two linear equations to solve for u' and v'. (That's where the Wronskian comes in.)

If we multiply the second equation by 2t^3, and subtract from the first equation, we get t^6v'= -12t- 20 or v'= -12t^{-5}- 20t^{-6}. If, instead, we multiply the second equation by 3t^3, and subtract from the first equation, we get -t^5u'= -12t- 20 or u'= 12t^{-4}+ 20t^{-5}.

Integrating those will give u and v to put into the original form.

 

[sunrah]

I would not call it "variation of constants"

Thanks this is quite different from how it was explained to us. BTW I study in German and here it is called "Variation der Konstanten" but I see your point