find the general solution of the non-homogeneous differential equation

Question

3f''+f'-2f = 4 +2x + e^x

Answer 1

Let f(x)=f₁(x)+f₂(x) and 3f₁"+f₁'-2f₁=0, 3f₂"+f₂'-2f₂=4+2x+eˣ.

To solve for f₁ we find the characteristic equation by solving:

3r²+r-2=(3r-2)(r+1), so r=⅔ and -1, and f₁=Ae^⅔x+Be^-x, where A and B are constants.

Now suppose f₂=a+bx+ceˣ, f₂'=b+ceˣ, f₂"=ceˣ, where a, b, c are constants.

3f₂"+f₂'-2f₂=3ceˣ+b+ceˣ-2a-2bx-2ceˣ. Matching coefficients, 3c+c-2c=1 (eˣ coefficients), so 2c=1, c=½.

-2b=2 (x coefficients) so b=-1.

b-2a=4 (constant term), so a=-5/2.

f₂=-5/2-x+½eˣ.

Therefore f(x)=(Ae^⅔x)+(Be^-x)-5/2-x+½eˣ.