I am wondering how to go about solving these differential equations. They are in the description. I only need one to be solved to learn how to solve the others, but tips for the others would be nice.

Question

Find the exact solution to the following differential equations. (I only need 1 explained/solved)

y'=(3-x)/(xy), x>0, y(1)=2

y'=y, y(2)=3

dy/dx+xy=x, y(0)=1/2

Answer 1

These all belong to a class of differential equations called "separable ordinary differential equations (ODE's)." It helps to write the y' as dy/dx. From there, you can treat dy/dx as a fraction of differentials: for instance in the third eq., you have

dy/dx+xy=x

dy/dx = x-xy = x(1-y)

dy=x(1-y)dx

1/(1-y) dy = x dx

Now, you can integrate both sides to obtain your answer. Recall that when we take indefinite integrals, we wind up with a +C term. The y(0)=1/2 bit is called the "initial condition." You would plug this point in (in this case (x,y)=(0,1/2)) to solve for C.

(Apologies for the awful formatting, Wyzant does not yet support LaTeX or similar formats in these response boxes.)