Calculus: Solve for y if y'= 1-4y; y=3 x=0?

Question

Please help me solve this problem showing both the method and the answer. Any helpful hints are welcome too.

Answer 1

Divide both sides by (1-4y) -- then you have

y'/(1-4y) = 1

Do a u substitution with u = (1-4y)
u' = -4y' -- so
y' = u'/(-4)

Now you have

(-1/4)*(u'/u) = 1

Take the antiderivative of both sides:

(-1/4)ln[u] = u+c

You can now put it in terms of "y" and enter 3. There should only be one value of C where it is an equality. There is no value of "x" -- so the fact that x = 0 is not relevant. Are you sure there is no "x"? Did you give us the right equation?

Answer 2

If it is -(3/4)x + 4 then that's a line. A line has area and variety both because the entire set of authentic numbers. you should use any x fee, and also you'll receive any y fee. If it is -3/(4x) + 4, then it is variety of trickier. you may't use x=0, because you may't divide through 0. the different style works in spite of the actuality that. So the area is the set of authentic numbers with the exception of 0. the range right that is problematic to discover. What you may do is come to a call for x and determine out what values of y can't be used: y = -3/(4x) + 4 y-4 = -3/(4x) (4x) (y-4) = -3 x (4y-16) = -3 x = -3 / (4y-16) the cost of y that would not artwork is even as 4y-16 = 0. which means the range is each and everything except y = 4. it is the set of all authentic numbers with the exception of four. The third chance, that can be the worst of all, is that if y = -3/(4x+4). the reason I say "worst" is because what you typed is not in any respect structure or form a similar as this. those parentheses are surely required, that is uncomplicated order-of-operations stuff. yet enable's attempt that one. obviously, the area is got here across like earlier: 4x+4 ? 0 4x ? -4 x ? -a million And back, we come to a call for the inverse and discover its area, to get the range of the unique: y = -3/(4x+4) 4x+4 = -3/y we may be able to stop right here. all of us understand already that y ? 0 is the really difficulty y-fee. So the range of your function is all numbers except 0. The area is all numbers except -a million.

Answer 3

y' = 1 - 4y
y'/(1 - 4y) = 1
ie dy/(1 - 4y) = dx
So ∫dy/(1 - 4y) = ∫dx
ie -¼ ln(1 - 4y) = x + c
So ln(1 - 4y) = -4x + k
So 1 - 4y = e^(-4x + k)
= Ae^(-4x) Where A = e^k)
So 4y = 1 - Ae^(-4x)
y = ¼(1 - Ae^(-4x)) There is NO real solution to the condition y = 3 when x = 0 as when x = 0 y = 0 so A = any real number

However if A is complex (= Be^iθ)
then 3 = ¼(1 - Be^iθ)
So Be^iθ = -11 = 11 x -1 = 11e^(2n + 1)iπ n is an integer
So y = ¼(1 - e^(-4x + 11e^(2n + 1)iπ)) for any integer n