dy/dx+2y=2, if y(0)=0?

Question

Find the solution to the following linear non-homogeneous differential equation of the first order if y(0)=0.

Answer 1

dy/dx+2y=2, if y(0)=0

dy/dx=2-2y=2(1-y)
dy/(1-y)=2dx
integrate
-ln(1-y)=2x+C{C is a constant}
ln(1-y)= -2x-C
take'e'
1-y= e^(-2x-C)
y= 1-e^(-2x-C)
take initial conditions
0=1-e^(-C)
>>>>>C=0
therefore,y=1-e^(-2x)

check in original equation
dy/dx=2e^(-2x)
2y=2-2e^(-2x)
dy/dx+2-2e^(-2x)=2
solution agrees with original
equation,therefore it is
correct

Answer 2

dy/dx + 2y = 2
dy/dx = 2 - 2y
Integral of (1/2-2y)dy = Integral dx
[ln(2-2y)]/(-2) = x + c
ln(2-2y) = -2x + C

Since y(0)=0, then
ln2 = C

So,
ln(2-2y) = -2x + ln2
ln(2-2y) - ln2 = -2x
ln[(2-2y)/2] = -2x
ln[2(1-y)/2] = -2x
ln(1-y) = -2x
1 - y = e^(-2x)
1 - e^(-2x) = y

Answer 3

dy/dx=2(1-y)
dy/(1-y)= 2dx
integrate both sides
ln(1-y) =-2x+const
or e^-(2x+const)=1-y
> y=1-e^(-2x) if const of int=0 since 1-e^0e^const=0
Missed neg sign!

Answer 4

Still doesn't make any sense.

Answer 5

dy/(1-y) = 2dx
-log(1-y)=2x
1-y = e^(-2x)
y=1 - e^(-2x)

Answer 6

Where the fcuk did "e" come from then ?

Answer 7

i c u p o k