How to solve $y''' – y = 2\sin(x)$

$$y”’ – y = 2\sin(x)$$

I’m doing differential equations and know pretty much all methods of solving them, but I haven’t come across anything of a higher order than second yet.

How do I go about solving this?

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Instead of solving the given differential equation, I’ll teach you how to fish.

Given $n\in \mathbb N$, given $a_0, \ldots ,a_n, \alpha, \beta\in \mathbb R$ and given the ODE
$$a_ny^{(n)}+a_{n-1}y^{(n-1)}+\ldots + a_0y=f \tag{ODE}$$

if $\forall x\in \mathbb R\left(f(x)=P(x)e^{\alpha x}\cos(\beta x)+Q(x)e^{\alpha x}\sin(\beta x)\right)$, for some polynomials $P$ and $Q$, then a particular solution $y_p$ of $\text{ODE}$ is determined by
$$\forall x\in \mathbb R\left[y_p(x)=x^k\left(R(x)e^{\alpha x}\cos(\beta x)+S(x)e^{\alpha x}\sin(\beta x)\right)\right],$$
where $R,S$ are polynomials such that $\deg\left(R\right)=\deg\left(S\right)=\max\left(\deg\left(P\right), \deg\left(Q\right)\right)$ and $k$ is the multiplicity of $\alpha +\beta i$ as a root of the characteristic polynomial of the homogeneous equation associated with $\text{ODE}$ ($a_n\lambda ^n+\ldots +a_1\lambda + a_0$), with the convention that $k=0$ if $\alpha +\beta i$ isn’t a root of the polynomial.

Adding your favorite solution $y_h$ of the homogeneous equation, yields the family of solutions $y_h+y_p$.

Example: Consider the differential equation determined by $$y”(x)-y'(x)+9y(x)=3\sin(3x)\tag{EX}$$

In the notation above $n=2, a_2=1, a_1=-1, a_0=9, \alpha=0, \beta=3, P(x)=0, Q(x)=3, f(x)=3\sin(3x)$ and $k=0$ (since $3i$ is not a root of $\lambda ^2-\lambda +9$).

So a particular solution to $\text{EX}$ is determined by $$y_p(x)=x^0\left[R(x)\cos(3x)+S(x)\sin(3x)\right] \tag{PS}$$
where $R$ and $S$ are polynomials whose degree is $0$, that is, they are constants, so for some $A,B\in \mathbb R$, $\text{PS}$ is equivalent to $$y_p(x)=A\cos(3x)+B\sin(3x).$$

Now replace the expression on the RHS of the above formula in $\text{EX}$.

One has $$y’_p(x)=-3A\sin(3x)+3B\cos(3x),$$ $$y”_p(x)=-9A\cos(3x)-9B\sin(3x).$$

Therefore, substituting in the equation and rearranging yields

Since $\{\cos , \sin\}$ is a linearly indepedent set over $\mathbb R$, it follows that $-3B=0$ and $3A=3$, yielding $y_p(x)=\cos(x)$.

Solving the homogeneous equation
should pose no problem: the roots of the characteristic polynomial $X^3-X$ are $0$, $-1$ and $1$. For a particular solution, consider that the derivative of a function of the form
g(x)=a\cos x+b\sin x
is a function of the same form.

One way to look for particular solutions to the given equation (I’ll assume you’re fine with the homogeneous solution, which is generally easier, in principle), is the following.

Consider the collection of function $f, f’, f”, f”’, f^{(iv)}, \ldots$. If at some point these functions form a finite-dimensional vector space i.e. there ends up being some linear relation of the form
f^{(n)} = \sum_{k=0}^{n-1}a_kf^{(k)}
then you can use as a guess for your particular solution the test function
If you substitute this into the differential equation, then you will find some linear equations which will determine the coefficients $A_k$. This is, in a nutshell, the method of undetermined coefficients.

In your case, $f = 2\sin(x)$. Its derivatives are $f’ = 2\cos(x)$ and $f” = -2\sin(x)$, which yields the relation
f” = (-1)f + 0 f’
and so we would use the guess
y_p(x) = Af(x) + Bf'(x)
or more simply,
y_p(x) = A\sin(x) + B\cos(x)

See if you can find a solution for
$$y”’-y=ae^{bx} $$
(Hint: try something $e$xponential). Then recall how sine (and cosine) and exponential are connected