Verified answer

y′ = 2(4 − y) and y(1) = 10

y'/(4 - y) = 2

∫dy/dx/(4 - y) dx = ∫ 2 dx

-ln(4 - y) = 2x + c

ln(4 - y) = -2x + c

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

y = ce^(-2x) + 4

10 = ce^(-2) + 4

c/e^2 = 6

c = 6e^2

y = 6e^2 * e^(-2x) + 4

y = 6e^(2 - 2x) + 4

cannot tell what your background is in math.....if you have any DE then you would know that y = c1 e^(-2t) + 4 ===> c1 = 6 e² ===> y(0) = 6 e² + 4...the line y = 4 is an asymptote ....if y(0) > 4 then curve decreases to the line while if y(0) < 4 then it increases to the line asymptote.

y' = 2(4 - y) =>

dy/(4 - y) = 2 dx =>

ln(4 - y) = -2x =>

4 - y = ce^(-2x).

If y(1) = 10, then

-6 = ce^(-2) and c = -6e^2,

so

4 - y = -6e^2 * e^(-2x).

Hence, when x = 0, you have

4 - y = -6e^2, and y = 4 + 6e^2.

Answer: y(0) = 4 + 6e^2.