Question

if dy/dx = 1+y, and when x=0, y=2, show that
y = 3e^x -1

Answer 1

use separation of variables
cross multiply
\[dy=(1+y)dx\]
divide both sides by 1+y
\[\frac{1}{1+y}dy=dx\]
integrate both sides
\[\int\limits_{}^{}\frac{1}{1+y}dy=\int\limits_{}^{}dx\]
\[\ln|y+1|=x+C\]
\[\ln|2+1|=0+C\]
\[C=\ln(3)\]
\[\ln|y+1|=x+\ln(3)\]
lets solve this for y
\[e^{\ln|y+1|}=e^{x+\ln3}\]
\[y+1=e^x+e^{\ln3}\]
\[y=e^x+3-1=e^x+2\]

Answer 2

oops made an error
e^(x+ln3) does not equal e^x+e^ln3 lol

Answer 3

\[y+1=e^x*e^{\ln3}\]
\[y+1=e^x*(3)\]
\[y=3e^x-1\]

Answer 4

thank you math genius!

Answer 5

btw where did x+c come from?

Answer 6

after you integrate you add constant

Answer 7

integrate what?

Answer 8

we integrated both sides
the x came from integrating 1 with respect to x

Answer 9

i got all my x's on one side and all my y's on the other side
then i integrated to find solve both sides

Answer 10

y+1=e^x*e^{ln3} ?

Answer 11

i don't get that part as well

Answer 12

law of exponents

Answer 13

x^(a+b)=x^a*x^b

Answer 14

y+1=e^x*(3) how did e^{ln3} become (3)?
and what happened to the absolute value brackets around y+1