Do you not have a teacher or at least a text book? There is a pretty standard formula for converting any linear first order differential equation to a single derivative using an "integrating factor". Given \(\displaystyle \frac{dy}{dx}+ f(x)y= g(x)\), multiply the equation by the "integrating factor" \(\displaystyle \mu(x)\): \(\displaystyle \mu \frac{dy}{dx}+ \mu f y= \mu g\).
A single derivative would be of the form \(\displaystyle \frac{d(\mu y)}{dx}\) which, by the product rule, is equal to \(\displaystyle \mu \frac{dy}{dx}+ \frac{d\mu}{dx}y\). Comparing that to \(\displaystyle \mu \frac{dy}{dx}+ \mu f y\), we see that we must have \(\displaystyle \frac{d\mu}{dx}= \mu f\). Solve that differential equation (which is a separable equation) for \(\displaystyle \mu\).
In fact, for any f(x) we can write that as \(\displaystyle \frac{d\mu}{\mu}= f(x)dx\) and, integrating \(\displaystyle ln(\mu)= \int f(x)dx+ C\). Taking the exponential, \(\displaystyle \mu(x)= C'e^{\int f(x)dx}\). Since we only need an integrating factor, we can take C' to be 1.
With this linear first order differential equation, \(\displaystyle \frac{dy}{dx}+ 3y= e^{2x}\), f(x) is the constant 3 so the integrating factor is \(\displaystyle \mu(x)= e^{\int 3dx}= e^{3x}\).
Multiplying on both sides of the equation by that \(\displaystyle e^{3x}\frac{dy}{dx}+ 3e^{3x}y= \frac{d(e^{3x}y)}{dx}= e^{5x}\). Integrate \(\displaystyle d(e^{3x}y)= e^{5x}dx\).
Subhotosh Kahn is suggesting a different method that applies to linear equations of any order, 1 or higher. A linear equation of order "n" would be an equation where the function, y, and its derivatives up to nth derivative are multiplied by functions of x and added. It is "homogeneous" if that is equal to 0, "non-homogeneous" if not. The idea is that we can find the general solution to a non-homogeneous equation by finding the general solution to the corresponding homogeneous equation (drop the function of x that is not multiplied by y or a derivative of y) then add to that any one solution to the entire equation.
Here the differential equation is \(\displaystyle \frac{dy}{dx}+ 3y= e^{2x}\) is "non-homogeneous" because of the "\(\displaystyle e^{2x}\)". Dropping that gives the "homogeneous" equation \(\displaystyle \frac{dy}{dx}+ 3y= 0\). That is the same as \(\displaystyle \frac{dy}{dx}= -3y\). We can separate that as \(\displaystyle \frac{dy}{y}= -3dx\) and integrate. To find a single solution to the entire equation, seeing that the "non-homogeneous part" is \(\displaystyle e^{2x}\), I would recommend setting \(\displaystyle y= Ae^{2x}\), for A a constant, putting it into the equation, and seeing what A must be in order to satisfy the equation.
