HELP ASAP

if dy/dx = 2 - y, and if y = 1 when x = 1, then y = ??? PHOTO ATTACHED

y=2-e^(1-x)

Step-by-step explanation:

This can be solved by separating the variables.

dy/dx=2-y

Multiply both sides by dx

Divide 2-y on both sides

dy/(2-y)=dx

Integrate both sides

-ln|2-y|=x+c

Multiply both sides by -1

ln|2-y|=-x-c

Equivalent logarithm form

2-y=e^(-x-c)

Minus 2 on both sides

-y=e^(-x-c)-2

Multiply both sides by -1

y=-e^(-x-c)+2

The ordered pair (1,1) belongs to the curve so we have

(I'm going to use this earlier form -ln|2-y|=x+c)

-ln|2-1|=1+c

-ln|1|=1+c

-0=1+c

0=1+c

Minus 1 on both sides

-1=c

y=-e^(-x-(-1))+2

Or

y=-e^(-x+1)+2

Or

y=2-e^(1-x)

this is not the answer this is just a question. is this middle school math?

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11