Question

Can someone show me the steps to solve these differential equations for y please: y'=2-y and y'=y(y+3)

Answer 1

y' = -y +2 separate into 2 parts:
part1 : y' = y, \(\dfrac{dy}{dx}= y\rightarrow \dfrac{dy}{y}= dx\) take integral both sides, you get |ln y|= x +C, hence \(y = e^{x+C}= e^x*e^C = Ce^x\)
part 2: y' =2, you have formula for this , that is if the right hand side is a constant, then \(y_{partial} = A\)
\(y'_{partial}= 0\) hence for partial y' +y =2, that is 0+A =2, then A =2 or \(y_{partial}=2\)
Now combine: general solution = y + y(partial)
\(y = Ce^x+2\)

Answer 2

problem2: y'= y (y+3)
\(\dfrac{y'}{y(y+3)}=1\\\dfrac{dy/dx}{y(y+3)}=1\\\dfrac{dy}{y(y+3)}=dx\)
integral both sides, the right hand side = x + C , right? hehehe easy, right?

Answer 3

The left hand side:
use partial decompose to solve it, that is
take the integrand out:
\(\dfrac{1}{y(y+3)}=\dfrac{A}{y}+\dfrac{B}{y+3}\)
hence A(y+3) + By =1
if y =0 , then 3A =1 , A = 1/3
if y =-3, then -3B = 1, B = -1/3
now, plug back to integral of the left hand side

Answer 4

\[\int\dfrac{dy}{y(y+3)}=\int (\dfrac{1}{3y}-\dfrac{1}{3(y+3)})dy\] just take term by term, you get it is = \(1/3 (ln y-ln(y+3))= 1/3 ln\dfrac{y}{y+3}\)
then let it = the right hand side which is x + C
I think you can handle the leftover, right?