Originally Posted by
MarkFL Let's look at the first one:
(1) $\displaystyle y'''+y'=4\sin(x)+2x^2$
The differential operator $\displaystyle D^2+1$ annihilates $\displaystyle \4\sin(x)$ and $\displaystyle D^3$ annihilates $\displaystyle 2x^2$ hence the operator:
$\displaystyle A\equiv D^3(D^2+1)$
annihilates $\displaystyle 4\sin(x)+2x^2$.
Thus, applying $\displaystyle A$ to both sides of (1) gives us:
(2) $\displaystyle D^4(D^2+1)^2[y]=0$
The characteristic roots are then:
$\displaystyle r=0,\,\pm i$
where $\displaystyle r=0$ is of multiplicity 4 and $\displaystyle r=\pm i$ are of multiplicity 2, and so the general solution to (2) is given by:
(3) $\displaystyle y(x)=c_1+c_2x+c_3x^2+c_4x^3+(c_5+c_6x)\cos(x)+(c_7 +c_8x)\sin(x)$
Now, recall that a general solution to (1) is of the form $\displaystyle y(x)=y_h(x)+y_p(x)$. Since every solution to (1) is also a solution to (2), then $\displaystyle y(x)$ must have the form displayed on the right-hand side of (3). But we recognize that:
$\displaystyle y_h(x)=c_1+c_2\cos(x)+c_3\sin(x)$
and so there must exist a particular solution of the form:
$\displaystyle y_p(x)=c_2x+c_3x^2+c_4x^3+c_6x\cos(x)+c_8x\sin(x)$
Now it is a matter of using the method of undetermined coefficients to determine the particular solution that satisfies (1).
So, what you need to do is compute $\displaystyle y_p'(x)$ and $\displaystyle y_p'''(x)$, substitute these into (1), and solve the resulting linear system that arises when you equate coefficients. Can you proceed?