Substituting w=y' in the Differential Equation

[Turion]

[tex]y'''-5y''+6y'=8+2sinx[/tex]

If I let w=y', would I be able to solve this differential equation? I'm currently stuck and I just want to know if making this substitution is why I am stuck.

 

[SteamKing]

Have you followed the procedure for solving a linear ODE?

First, solve the homogeneous equation.
Second, find the particular solution.
Third, add the particular solution to the complementary solution (of the homogeneous equation).

 

[arildno]

[tex]y'''-5y''+6y'=8+2sinx[/tex]

If I let w=y', would I be able to solve this differential equation? I'm currently stuck and I just want to know if making this substitution is why I am stuck.

Yes, that works nicely.
If we look at the homogenous system, you get the characteristic polynomial, with e^rx as trial solution:
r^2-5r+6=0, giving r=3 and r=2 as possibles. Furthermore, you have w_p1=4/3, for the constant particular solution.
Now, in order to solve for the particular solution 2sinx, you generally will need w_p=Asin(x)+Bcos(x)

Inserting this gives you the two equations in A and B

sin(x): -A+5B+6A=2 goes to: 5A+5B=2
cos(x):-B-5A+6B=0 goes to: 5B-5A=0
Thus, A=B=1/5.