Is every linear differential equation solvable using the standard method? I am asking this question because I was not able to find the solution to this linear differential equation.
$$\dfrac {dy}{dx} + 3xy = \sin(x) $$
 A: According to Wolfram Alpha, the solution is $$\DeclareMathOperator{\erfi}{erfi}y(x) = c_1\cdot e^{-(3x^2)/2} + \frac{1}{2}\cdot i\sqrt{π/6}\cdot e^{1/6-(3 x^2)/2}\erfi(\frac{3x-i}{\sqrt{6}})-\frac{1}{2}i\sqrt{π/6}\cdot e^{\frac{1}{6}-(3 x^2)/2}\erfi({3x+i}{\sqrt{6}})$$
Which is an abominable mess. Linear ordinary differential equations are solvable, but always run the risk of being incredibly difficult! This isn't really an answer, but the solution just wouldn't fit into the comments.
If you're curious, "erfi" is the imaginary error function.
Correction: to the best of my understanding, a linear o.d.e of the form $y'(x)+Py=Q$, where $R,Q$ are strictly functions of $x$ only, is solvable if the integral $\int RQ\,dx$ is solvable, where $R$ is $e^{\int P\,dx}$ such that $Ry'(x)+RPy=RQ\implies\frac{d}{dx}[Ry]=RQ$. Here, it was solvable since the ridiculous integral of "RQ" (i.e. $\sin(x)e^{3x^2/2}$) was technically solvable by a computer.
Perhaps there are other forms of linear o.d.e that are not, in general, solvable, but I've yet to see them; hence why this is not really an answer, just a long comment.
A: $$\dfrac {dy}{dx} + 3xy = 0 \implies y=c\,e^{-\frac{3 x^2}{2}}$$ Variation of parameter leads to
$$\dfrac {dc}{dx}=\sin(x)\,e^{\frac{3 x^2}{2}}\implies c=\int  \sin(x)\,e^{\frac{3 x^2}{2}}\,dx + C$$
Just stay with that. I do not suppose that they expect you to know about the imaginary error function with, moreover, complex arguments.
