How to solve this Diff Eq (with multiple terms) $$\frac{dy}{dx}=7xy$$
I know this turns into
$$\frac{dy}{y}=7xdx$$
.....etc.
But, how do you solve the following:
$$\frac{dy}{dx}=7x+y$$
Not sure how to seperate the parts to respective sides.
 A: You're right, it can't be done using seperation of variables.  Here's one way you can do it:  \begin{array}{l} {\frac{dy}{dx} =7x+y} \\ {\frac{dy}{dx} -y=7x} \\ {dy-ydx=7xdx} \\ {-\left(7x+y\right)dx+dy=0} \\ {-e^{-x} \left(7x+y\right)dx+e^{-x} dy=0} \\ {\frac{\partial }{\partial y} \left(-e^{-x} \left(7x+y\right)\right)=-e^{-x} } \\ {\frac{\partial }{\partial x} \left(e^{-x} \right)=-e^{-x} } \\ {\frac{\partial F}{\partial x} =-e^{-x} \left(7x+y\right)} \\ {\frac{\partial F}{\partial y} =e^{-x} } \\ {F\left(x,y\right)=e^{-x} y+G\left(x\right)} \\ {\frac{\partial F}{\partial x} =-e^{-x} y+\frac{dG}{dx} =-e^{-x} \left(7x+y\right)} 
\\ {\frac{dG}{dx} =-7xe^{-x} } 
\\ {dG=-7xe^{-x} dx=7xd\left(e^{-x} \right)=d\left(7xe^{-x} \right)-e^{-x} d\left(7x\right)}
\\ {=d\left(7xe^{-x} \right)-7e^{-x} dx=d\left(7xe^{-x} \right)+d\left(7e^{-x} \right)=d\left(7e^{-x} \left(x+1\right)\right)} 
\\ {G\left(x\right)=7e^{-x} \left(x+1\right)+D} 
\\ {F\left(x,y\right)=e^{-x} y+7e^{-x} \left(x+1\right)=C} \\ {y+7\left(x+1\right)=Ce^{x} } \\ {y=Ce^{x} -7\left(x+1\right)} \end{array}
To summarize, what I did is put the equation in the form $M(x,y)dx+N(x,y)dy=0$, multiplied both sides by the integrating factor $e^{-x}$, verified that we had an exact differential equation of the form $dF=0$ for some $F(x,y)$, solved a pair of partial differential equations to find what $F(x,y)$ was, and then solved $F(x,y)=C$ to find $y$.  Along the way I used partial differentiation, partial integration, and integration by parts.
A lot of this may look like gibberish to you since you're just starting calculus, and there are easier ways of doing this particular problem (since it's just a linear equation with constant coefficients), but I just wanted to give you a glimpse of the beautiful mathematical techniques that await you if you continue with multivariable calulus and differential equations.
A: Here is another approach (guessing, otherwise known as undetermined coefficients).
We call this part the complementary solution, we have 
$$y' - y = 0$$
Do you  know a function that will provide a zero when differentiated and subtracted from itself?
Choose $y_c(x) = c e^x$.
We now have what is called the particular part of $y' - y = 7x$
Lets guess at $y_p = ax +b$ and substitute back in:
$$y' - y = a - (a x + b) = 7 x \implies a = b = -7$$
The final solution is the linear combination of complementary and particular, hence:
$$y(x) = y_c(x) + y_p(x) = ce^x -7x -7$$
As a check, substitute back into ODE and see you get equality.
We could have used various methods to solve this that include integrating factor, linear substitution, Laplace Transform, exact equation, series methods, Picard Iteration...
