Question

How can I solve this differential equation (y'''-6y''=3-cos x) by undetermined coefficients?

`y'''-6y''=3-cos x`

Answer 1

`y=Ae^(6x)+Bx+C-1/4x^2+1/37sinx-6/37cosx`

Explanation:

Characteristic equation of this differential one: `r^3-6r^2=0` or `r^2*(r-6)=0`

Hence roots of it `r_1=r_2=0` and `r3=6`

Consequently, homogeneous part of solution is `y_h=Ae^(6x)+Bx+C`

Due to `Bx` and `C` have part of homogenous solution, particular solution must be in form: `y_p=Dx^2+Esinx+Fcosx`

Hence,

`y'=2Dx+Ecosx-Fsinx`, `y''=2D-Esinx-Fcosx` and `y'''=-Ecosx+Fsinx`

`(-Ecosx+Fsinx)-6*(2D-Esinx-Fcosx)=3-cosx`

`-12D+(F+6E)*sinx+(-E+6F)*cosx=3-cosx`

After equating coefficients,

`-12D=3`, `F+6E=0` and `-E+6F=-1`

After solving these equations, I found

`D=-1/4`, `E=1/37` and `F=-6/37`

Consequently, `y_p=-1/4x^2+1/37sinx-6/37cosx`

Thus,

`y=y_h+y_p=Ae^(6x)+Bx+C-1/4x^2+1/37sinx-6/37cosx`