Differential equation with a unknown function $g$ but a known solution. Say we have an equation $y'+g(x)y=3x, x>0$.


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*We know one solution to this FDE: $y(x) = x^2$


How can we know if there's a solution which satisfies $y(1) = 2$?

Questions I'd like answered


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*Q1: First and foremost, how do I go about solving the problem I have listed?

*Q2: If I understand correctly, this is a first-order linear ordinary differential equation. However, is there any known term for the problem I have, where you have an unknown function $g(x)$.

*Q3: Is $g(x)$ to be considered a coefficient for our differential equation (where we are searching for $y$).

A1: For this type of problem, you have for a fact that one solution is $y(x) = x^2$ which gives you that $\frac{dy}{dx}=2x$. Using this in your equation, you can solve for $g(x)$.
$$2x+g(x)\cdot x^2 = 3x \iff g(x) = \frac{1}{x}$$
Now, using $g(x)$, you solve for $y$ by the method you used.
$$y' + \frac{y}{x}=3x$$
Multiply by $e^{\int g(x)} = e^{\ln x} = x$:
$$\begin{eqnarray}
xy'+y&=&3x^2 \\
D(xy) &=& 3x^2 \\
xy &=& \int 3x^2 = 3\int x^2 = x^3 + C \\
y &=& x^2 + \frac{C}{x}
\end{eqnarray}$$
In the question, we had the initial value $y(1) = 2$, and as such we get:
$$2 = 1^2+\frac{C}{1} = 1+C \iff C = 1$$
The sought-after function is $\boxed{y(x) = x^2+ \frac{1}{x}}$
A2: Couldn't find any name for it.
A3: Still don't know.
