Variation of Parameters

• hils0005
In summary, the conversation discusses the process of solving the equation y^(4)-6y^(3)=-5sinx using the variation of parameters method. The solution involves finding the roots of the characteristic equation, using a particular solution with sine and cosine functions, and combining like functions to solve for the coefficients. The final general solution is given as y=A + Be^6x + Dx + Ex^2 + 5/37sinx -30/37cosx.

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?

hils0005 said:

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 "1295Be6x" and -6(216Be6x)" 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.

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

hils0005 said:
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.

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??

-5/37sinx +30/37cosx

1. What is the concept of "Variation of Parameters"?

Variation of Parameters is a method used in solving differential equations by finding the particular solution using a linear combination of the solutions to the corresponding homogeneous equation.

2. When is the "Variation of Parameters" method used?

This method is typically used when the coefficients of the differential equation are not constant, and the right-hand side of the equation is a non-zero function.

3. How is the "Variation of Parameters" method applied?

The first step is to find the solutions to the corresponding homogeneous equation. Then, a formula is used to find the particular solution by integrating a product of the solutions and the non-homogeneous function.

4. What are the advantages of using the "Variation of Parameters" method?

One advantage is that it can be used for a wide range of differential equations, including those with non-constant coefficients. It also does not require any initial conditions to be known.

5. Are there any limitations to the "Variation of Parameters" method?

One limitation is that it can be more complex and time-consuming compared to other methods. It also may not always be applicable to higher-order differential equations or systems of differential equations.