differential equation 2?

Question

if dy/dx = 10 - (y/10)
and y = 10 when x = 0
find y in terms of x

i tried solving the question and i got the same answer
but the book's answer is different
its:
y = 90e^(-x/10) -100
(opposite signs)

Answer 1

dy/dx = 10 - (y/10)

dy / (10 - (y/10)) = dx .... Put the y terms on one side and the x on the other

Integrate.
dy / (10 - (y/10)) = 10 dy/(100 - y)
This is of the form du/u which integrates to ln(u)
And the integral of dx is just x

so -10ln(100 - y) = x + C ... You must toss in that constant

Solve for C using y =10 when x=0: -10ln(90) = C

Now put it into terms of y = a function of x
-10ln(100 - y) = x -10ln(90)
ln(100 - y) = (ln(90) - x/10)
100 - y = exp [ln(90) - x/10]
y = 100 - exp [ln(90) - x/10]

Differentiate to check:
dy/dx = d/dx[100 - exp [ln(90) - x/10]]
dy/dx = (1/10) exp [ln(90) - x/10]
dy/dx = (1/10) (100 - y) = 10 - y/10

Answer 2

This is a first order linear diff equation. Solve it by adding y/10 to each side, then multiplying thru by an integrating factor: e^int(1/10). The equation is then of a form:
d[e^(1/10 x) * y] dx = 10 e^(1/10 x). Integrate both sides and you get:

e^(1/10 x) * y = 100e^1/10 x + C, use the initial condition y = 10 when x = 0, and I get C = - 90.

Hope you can read that, it's hard to write calculus in a text form!

Answer 3

dy/dx = 10 - y/10 = (100 - y).10
1/(100-y)dy = 1/10 dx

Integrating both sides
- ln(100 - y) = (x + a)/10

Substituting x = 0 and y =10
- ln 90 = a/10
a = - 10 ln90

Substitute value of a
- ln(100 - y) = (x - 10 ln90)/10 = x/10 - ln90
ln(100 - y) = -x/10 + ln90
100 - y = e^(-x/10 + ln90)
y = 100 - e^(-x/10 + ln90)

Answer 4

desert eagle. You have done much harder questions than this one. You can certainly do this one. This is a separable differential equation. Get the y's on one side and x's on other and integrate.
Alternatively, you can write it as dy/dx + (y/10) = 10
and multiply by integrating factor: v(x) = e^(integral 1/10 dx)
and y = 1/v(x) Int ( v(x) * 10)dx.

Answer 5

dy/dx = (1/10)(100-y)
dy/(100-y) =dx/10
Integrating both sides gives,
-ln|100-y| + ln90 = x/10
100-y = e^(ln90 - x/10)
y = 100 - e^(ln90 - x/10)
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Ideas: Using definite integral: x from 0 to x, y from 10 to y can avoid solving for the constant of integral, and thus simplifies the process.