Solve the differential equation
$$y''+8y'+32y=8x^{3}e^{-4x}\sin(4x)$$
Using the method of undetermined coefficients
What i tried:
The auxiliary equation corresponding to that equation is
$$r^2+8r+32=0\Longrightarrow r=\frac{-8\pm\sqrt{64-128}}{2}=\frac{-8\pm 8i}{2}=-4\pm 4i$$
So $$y_{c}=A_{1}e^{(-4+4i)x}+A_{2}e^{(-4-4i)x}=c_{1}e^{-4x}\cos(4x)+c_{2}e^{-4x}\sin(4x)$$
But i did not understand how to find a particular solution using method of Undetermined Coefficients.
Please Help me. Thanks
Best Answer
For getting P.I.,
Just do , $$\frac{8\cdot e^{(4i-4)x}\cdot x^{3}}{D^{2}+8D+32} $$ $$=\frac{8\cdot e^{(4i-4)x}\cdot x^{3}}{(D+4)^{2}+16} $$ $$=8\cdot e^{(4i-4)x} \frac{x^{3}}{(D+4i-4+4)^{2}+16} $$ $$=8\cdot e^{(4i-4)x} \frac{x^{3}}{D^{2}+8Di} $$ $$=(-i)\cdot e^{(4i-4)x} (\frac{1}{D}-\frac{1}{8i}+\frac{D}{(8i)^{2}}-\frac{D^2}{(8i)^3}+\frac{D^3}{(8i)^4}-\frac{D^4}{(8i)^5}+\cdot\cdot\cdot)\cdot x^{3} $$ $$=(-i)\cdot e^{(4i-4)x} (\frac{x^4}{4}-\frac{x^3}{8i}+\frac{3x^2}{(8i)^{2}}-\frac{6x}{(8i)^3}+\frac{6}{(8i)^4}) $$ I think , you would get it from here,,, for the P.I. , just take the imaginary part .