Question

Find the solution of

(D^2-4D+4)y=12x+2e^(3x)

Answer 1

The given DE is $\displaystyle (D^2-4D+4)y=12x+2e^{3x}$.

Now, the auxiliary equation is;

$m^2-4m+4=0\Rightarrow(m-2)^2=0\Rightarrow m=2$ (twice)

$\therefore y_c=(c_1+c_2x)e^{2x}$, where $c_1,c_2$ are constants.

Next, to obtain the particular integral;

$y_p=(D^2-4D+4)^{-1}12x+\frac{1}{D^2-4D+4}2e^{3x}\\ \quad\ =3\left[1-D+\frac{D^2}{4}\right]^{-1}x+\frac{1}{9-12+4}2e^{3x}\\ \quad\ =3\left[1+D-\frac{D^2}{4}\right]x+2e^{3x}\\ \quad\ =3\left[x+D(x)-\frac{D^2}{4}(x)\right]+2e^{3x}\\ \quad\ =3[x+1-0]+2e^{3x}=3x+3+2e^{3x}$

Thus, the general solution is;

$y=y_c+y_p=(c_1+c_2x)e^{2x}+3x+3+2e^{3x}$