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Math Help - derivative problem

  1. #1
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    derivative problem

    if dy/dx =ysec^2x and y=5 when x=0 then y=

    a. e^tanx +4
    b. e^tanx +5
    c. 5e^tanx
    d. tanx +5
    e. tanx +5e^x
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  2. #2
    o_O
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    \frac{dy}{y} = sec^{2} x dx

    Integrate:
    \ln |y| = \tan x + C

    Plug in (0, 5):

    C = \ln 5

    \ln|y| = \tan x + \ln 5
    e^{\ln |y|} = e^{ \tan x + \ln 5} (Express both sides as powers of e)

    Can you go on from here?

    Remember: e^{\ln(x)} = x, y = e^{x} > 0 and a^{m + n} = a^{m} a^{n}
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  3. #3
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    Hello, sandiego234!

    If \frac{dy}{dx} \:=\:y\sec^2\!x\;\;\text{ and }y=5 \text{ when }x=0, then y \:=

    a)\;e^{\tan x} +4\qquad b)\;e^{\tan x} +5\qquad c)\;5e^{\tan x}\qquad d)\;\tan x +5\qquad e)\;\tan x +5e^x
    This is a differential equation . . .


    Separate variables: . \frac{dy}{y} \;=\;\sec^2\!x\,dx

    Integrate: . \ln y \;=\;\tan x + C


    When x = 0,\:y=5\!:\;\;\ln 5 \;=\;\tan 0 + C\quad\Rightarrow\quad C = \ln5

    . . Hence: . \ln y \:=\:\tan x + \ln 5


    We have: . \ln y - \ln 5 \:=\:\tan x\quad\Rightarrow\quad \ln\left(\frac{y}{5}\right) \:=\:\tan x

    Therefore: . \frac{y}{5} \:=\: e^{\tan x}\quad\Rightarrow\quad \boxed{y \:=\:5e^{\tan x}}\quad\hdots \text{answer (c)}

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  4. #4
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    thank you soroban, we're both from MA...i'm over in Acton.

    I like your way of explaining. succinct.

    i'm aruban, because i lived there for a good 7 years
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