Y′′=−20⋅4x^3, Second order linear ordinary DE

[Feodalherren]

Homework Statement

y′′=−20⋅4x^3

Homework Equations

Undetermined coefficients method

The Attempt at a Solution

so at first, solving the associated homogeneous equation I find the fundamental set of solutions to be: y1=1 and y2=x.

I know that these are correct. Now for the part that confuses me.

I'm trying to find Yp, the particular solution.

g(x) = -80x^3 so Yp is of the form Ax^3 + Bx^2 + Cx + D

but when I solve this I get Yp=0, that is not correct.

It seems to me that Ax^3+Bx^2+Cx+D is Linearly independent from y1 and y2 so why doesn't this work? The book lists the solution as Yp=-4x^5.

 

[Mark44]

Homework Statement

y′′=−20⋅4x^3

Homework Equations

Undetermined coefficients method

The Attempt at a Solution

so at first, solving the associated homogeneous equation I find the fundamental set of solutions to be: y1=1 and y2=x.

This is much simpler than you're making it out to be.
Just integrate both sides twice.

I know that these are correct. Now for the part that confuses me.

I'm trying to find Yp, the particular solution.

g(x) = -80x^3 so Yp is of the form Ax^3 + Bx^2 + Cx + D

but when I solve this I get Yp=0, that is not correct.

It seems to me that Ax^3+Bx^2+Cx+D is Linearly independent from y1 and y2 so why doesn't this work? The book lists the solution as Yp=-4x^5.

 

[Feodalherren]

Ok fair enough, it could be done like that. But the point of this chapter is to learn this method. The question specifically asks for the fundamental set of solutions, Yp and the general solution. At any rate, I'm stuck on another similar problem which I can't integrate. I'm sure I can solve it if I can figure out this simpler case first.

 

[Mark44]

Ok fair enough, it could be done like that. But the point of this chapter is to learn this method. The question specifically asks for the fundamental set of solutions, Yp and the general solution. At any rate, I'm stuck on another similar problem which I can't integrate. I'm sure I can solve it if I can figure out this simpler case first.

Since y'' = -80x³, a reasonable choice for y_p would be $y_p = Ax^5 + Bx^4 + Cx^3 + Dx^2 + c_1x + c_2$. Differentiate twice and compare coefficients to what you have in the differential equation. That seems like a pointless technique, though, IMO, at least for this problem.

For the nonhomogeneous equation, y'' = 0, the fundamental set of solutions is as you say, {1, x}. Note that in my particular solution above, the complementary solutions are included with coefficients of c₁ and c₂.

BTW, is there some reason you wrote -20*4x³ instead of -80x³?

 

[Feodalherren]

Ok that works but how did you pick your Yp? I know Yp has to be linearly independent from the fundamental set of solutions but it seems to me like Ax^3+Bx^2+Cx+D already is, so why doesn't that satisfy the conditions? I feel like I'm missing something very fundamental here.

 

[SteamKing]

Ok that works but how did you pick your Yp? I know Yp has to be linearly independent from the fundamental set of solutions but it seems to me like Ax^3+Bx^2+Cx+D already is, so why doesn't that satisfy the conditions? I feel like I'm missing something very fundamental here.

If yp = Ax^3+Bx^2+Cx+D, what happens after you differentiate yp twice w.r.t x? Do you get a cubic in x?

 

[Feodalherren]

I get 6Ax + 2B

 

[Feodalherren]

So essentially what you're telling me that if g(x) is a polynomial of degree n, then for y'' the 2nd derivative of yp has the be a polynomial of degree n?

 

[Mark44]

So essentially what you're telling me that if g(x) is a polynomial of degree n, then for y'' the 2nd derivative of yp has the be a polynomial of degree n?

Something like that.

There is the Method of Annihilators, which is applicable in this problem.
Homogeneous equation: y'' = 0
The characteristic equation is r² = 0, which has r = 0 as a repeated root. The complementary function (solution of the homogeneous problem) is y_c = c₁ + c₂x.

The nonhomogeneous problem (your problem) is y'' = -80x³. This could also be written as D²y = -80x³, where Dy means dy/dx or y', and D²y means $\frac{d^2y}{dx^2}$ or y''.

We note that the D⁴ operator "annihilates" the right side, as D⁴(x³) = 0. IOW, the fourth derivative of x³ = 0. Because of this fact, we apply the D⁴ operator to both sides of the nonhomogeneous problem to get
D⁴(D²y) = D⁴(-80x³) = 0
Or, D⁶y = 0, which is a homogeneous equation.

The characteristic equation is r⁶ = 0, so that r = 0 is a repeated root. The general solution of this equation is $y = c_1 + c_2x + Ax^2 + Bx^3 + Cx^4 + Dx^5$. I wrote the first two differently, as they were the solution of y'' = 0, and I want to keep them separate. The particular solution is the portion with the coefficients A, B, C, and D.

 

[Feodalherren]

Perfect, that was exactly the kind of explanation that I was looking for. Thanks!