Solve Euler's Equation: x3y'''+15x2y''+61xy'+64y=0

[manenbu]

Given the homogeneous equation:
x³y''' + 15x²y'' + 61xy' + 64y = 0

I get 3 solutions which are all -4.

Does this mean that the solution for y should be:

y = c₁x^-4 + c₂x^-4lnx + c₃x^-4ln²x
?

 

[matematikawan]

I think you get it right.

 

[pbandjay]

If they are all solutions to the differential equation, then, by the superposition principle, the sum is a solution.

 

[HallsofIvy]

Notice, by the way, that the substitution t= ln(x) changes the differential equation into the "constant coefficients" equation [itex]d^3y/dt^3+ 12d^2y/dt^2+ 48dy/dt+ 64y= 0[/itex] which has characteristice equation [itex]r^3+ 12r^2+ 48r+ 64= (r+ 4)^2= 0[/itex] so the general solution to the constant coefficients equation is [itex]y(t)= C_1e^{-4t}+ C_2te^{-4t}+ C_3t^2 e^{-4t}[/itex] and the general solution to the original equation is [itex]y(x)= C_1e^{-4ln(x)}+ C_2 ln(x) e^{-4ln(x)}[/itex][itex]+ C_3 (ln(x))^2 e^{-4ln(x)}[/itex][itex]= C_1x^{-4}[/itex][itex]+ C_2 ln(x) x^{-4}+ C_3 (ln(x))^2 x^{-4}[/itex], just as you say.