How Do You Solve a Fourth Order Differential Equation with Sinusoidal Forcing?

[hils0005]

[SOLVED] Variation of Parameters

Homework Statement

y^(4)-6y^(3)=-5sinx

The Attempt at a Solution

I factored this at x^3(x-6)=0
so my r values are 0,6
also using for y(p) Dcosx + Esinx

y=Ae^0 + Be^6x + Dcosx + Esinx ?
y' =6Be^6x -Dsinx + Ecosx
y'' =36Be^6x-Dcosx - Esinx
y''' =216Be^6x + Dsinx - Ecosx
y'''' =1296Be^6x +Dcosx + Esinx

(1296Be^6x +Dcosx + Esinx) - 6(216Be^6x + Dsinx - Ecosx)=-5sinx

is this the correct way to set up this problem?

 

[HallsofIvy]

Homework Statement

y^(4)-6y^(3)=-5sinx

The Attempt at a Solution

I factored this at x^3(x-6)=0
so my r values are 0,6
also using for y(p) Dcosx + Esinx

y=Ae^0 + Be^6x + Dcosx + Esinx ?

This is a fourth order equation. The solution space of the homogeneous equation is a four[\b] dimensional vector space. You need 4 independent solutions to it and you only have 2. What do you do when you have a double or triple root to your characteristic equation?

y' =6Be^6x -Dsinx + Ecosx
y'' =36Be^6x-Dcosx - Esinx
y''' =216Be^6x + Dsinx - Ecosx
y'''' =1296Be^6x +Dcosx + Esinx

(1296Be^6x +Dcosx + Esinx) - 6(216Be^6x + Dsinx - Ecosx)=-5sinx

is this the correct way to set up this problem?

Obviously the "1295Be^6x" and -6(216Be^6x)" terms will cancel out. When solving for a "particular solution", there is no need to carry the solutions to the homogeneous equation along. Just use y= A sin(x)+ B cos(x) and then, after you have found a particular solution, add it to the homogeneous solution.

 

[hils0005]

you put in a faxtor of x and x^2 ?
y(c)= Ae^0 + Be^6x + Dxe^0 + Ex^2e^0
I used 0 because that would be consitent with the x^3 term-

y(c)=A + Be^6x + Dx + Ex^2

y(p)=Asin(x) + Bcos(x)
y'(p)=Acosx - Bsinx
y"(p)=-Asinx - Bcosx
y'''(p)=-Acosx + Bsinx
y""(p)=Asinx + Bcosx

(Asinx + Bcosx) - 6(-Acosx + Bsinx) = -5sinx

how can you solve this??
A(sinx + 6cosx) + B(cosx-6sinx)=-5sinx

 

[HallsofIvy]

you put in a faxtor of x and x^2 ?
y(c)= Ae^0 + Be^6x + Dxe^0 + Ex^2e^0
I used 0 because that would be consitent with the x^3 term-

y(c)=A + Be^6x + Dx + Ex^2

Yes, that is correct for the homogeous equation.

y(p)=Asin(x) + Bcos(x)
y'(p)=Acosx - Bsinx
y"(p)=-Asinx - Bcosx
y'''(p)=-Acosx + Bsinx
y""(p)=Asinx + Bcosx

(Asinx + Bcosx) - 6(-Acosx + Bsinx) = -5sinx

how can you solve this??
A(sinx + 6cosx) + B(cosx-6sinx)=-5sinx

Better is to "combine like functions": (A- 6B)sinx+ (6A+ B)cos x= -5sin x

Now, since sin x and cos x are "independent" functions, The coefficients of like functions, on opposite sides of the equation, must be equal. That gives you two equations to solve for A and B.

 

[hils0005]

two equations would then be:
1.(A-6B)sinx=-5sinx
2.(6A+B)cosx=0

1.(A-6B)sinx=sinx
A-6B=-5
-B/6 - 6B=-5
B=-30/37

6Acosx+Bcosx=0
6Acosx=-Bcosx
A=-B/6

A=-(-37/30)/6= 5/37

So general solution:
y=A + Be^6x + Dx + Ex^2 + 5/37sinx -30/37cosx

There seems like there is a different way to solve this??

 

[hils0005]

-5/37sinx +30/37cosx