$Y'''+Y''+3Y'-5Y=5\sin2x+10x^2+3x+7 ?$ I couldn't solve this one. A help will be appreciated. I guess transforming variables to m values by using Cauchy's method is very simple($m^3+m^2+3m-5$). After this point I can't put things together with right side of the equation. What should be the next step?
 A: First, let's find the solution to the homogeneous equation
$$ Y_{h}''' + Y_{h}'' + 3Y_{h}' - 5Y_{h} = 0.$$
Using the differential operator $D$, we have $(D^{3} + D^{2} + 3D - 5)Y_{h} = 0$. The equation $x^{3} + x^{2} + 3x - 5$ has a root $x=1$ (by inspection). So $x^{3} + x^{2} + 3x - 5 = (x-1)(x^{2}+2x+5)$. Hence, the other two roots are
$$ x= \frac{-2 \pm \sqrt{-16}}{2}= -1\pm 2i.$$
Using these roots, you should know how to find the homogeneous solution $Y_{h}$.
Now, let's find the solution to $Y_{1}''' + Y_{1}'' + 3Y_{1}' - 5Y_{1} = 5\sin(2x)$. You can guess that the solution will be of the form $Y_{1} = a \sin(2x) + b \cos(2x)$. If you plug this $Y_{1}$ into the differential equation at the beginning of this paragraph, you will find a system of linear equations involving $a$ and $b$ which will allow you to find the right coefficients.
Finally, find the solution to $Y_{2}'''+Y_{2}''+3Y_{2}'-5Y = 10x^{2}+3x+7$. This time, use the guess $Y_{2} = ax^{2} + bx + c$. Again, plug $Y_{2}$ into the differential equation, find a system of linear equations involving $a,b,c$, and solve to find the correct coefficients.
Your final answer will be $Y_{h} + Y_{1} + Y_{2}$.
