derivative problem

**sandiego234** · April 24th 2008, 03:58 PM

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

**o_O** · April 24th 2008, 04:11 PM

$\frac{dy}{y} = sec^{2} x dx$

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 $a^{m + n} = a^{m} a^{n}$

**Soroban** · April 24th 2008, 04:17 PM

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

**sandiego234** · April 24th 2008, 07:04 PM

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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