Answer to Question #150198 in Differential Equations for Hamza

Question

(D^3 - 27) = e^3x + e^-3x

Answer 1

$y^{(3)}-27y=e^{3x}+e^{-3x}$

First we solve the homogenous equation $y^{(3)}-27y=0$

The characterisic equation is $\lambda^3-27=0$, which has roots $\lambda_1=3, \lambda_2=3e^\frac{2i\pi}{3}=-\frac{3}{2}+\frac{3\sqrt{3}}{2}i, \lambda_3=3e^\frac{-2i\pi}{3}=-\frac{3}{2}-\frac{3\sqrt{3}}{2}i$

So the solution of $y^{(3)}-27y=0$ is $y=Ae^{3x}+Be^{-\frac{3}{2}x}\cos\frac{3\sqrt{3}}{2}x+Ce^{-\frac{3}{2}x}\sin\frac{3\sqrt{3}}{2}x$

Now we find partial solution of $y^{(3)}-27y=e^{3x}+e^{-3x}$ in form $y=Vxe^{3x}+We^{-3x}$

By the General Leibniz rule we have $(xe^{3x})^{(3)}=x^{(3)}e^{3x}+3x''(e^{3x})'+3x'(e^{3x})''+x(e^{3x})^{(3)}=$

$=0+0+3\cdot 9e^{3x}+27xe^{3x}=27e^{3x}+27xe^{3x}$

So $y^{(3)}-27y=27Ve^{3x}+27Vxe^{3x}-27We^{-3x}-$

$-27Vxe^{3x}-27We^{-3x}=27Ve^{3x}-54We^{-3x}$

Then $27V=-54W=1$, that is $V=\frac{1}{27},W=-\frac{1}{54}$

We obtain the general solution $y=Ae^{3x}+Be^{-\frac{3}{2}x}\cos\frac{3\sqrt{3}}{2}x+Ce^{-\frac{3}{2}x}\sin\frac{3\sqrt{3}}{2}x+$

$+\frac{1}{27}xe^{3x}-\frac{1}{54}e^{-3x}$

Answer: $Ae^{3x}+Be^{-\frac{3}{2}x}\cos\frac{3\sqrt{3}}{2}x+Ce^{-\frac{3}{2}x}\sin\frac{3\sqrt{3}}{2}x+$

$+\frac{1}{27}xe^{3x}-\frac{1}{54}e^{-3x}$