$\displaystyle dy/dx = -6xy$

$\displaystyle \int \frac{1}{y} dy = \int -6x dx$

$\displaystyle ln y = -3x^2 + c$

$\displaystyle y = e^{-3x^2 + c} $

is my answer Correct ?

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- Apr 24th 2013, 08:44 AMmastermin346differential equation need Help
$\displaystyle dy/dx = -6xy$

$\displaystyle \int \frac{1}{y} dy = \int -6x dx$

$\displaystyle ln y = -3x^2 + c$

$\displaystyle y = e^{-3x^2 + c} $

is my answer Correct ? - Apr 24th 2013, 08:49 AMBobPRe: differential equation need Help
Yes, though it can be taken one step further,

$\displaystyle y = e^{-3x^{2}+c}=e^{-3x^{2}}.e^{c}=Ae^{-3x^{2},$

which, subsequently, will be slightly easier to deal with. - Apr 24th 2013, 08:55 AMmastermin346Re: differential equation need Help
its given y(0) = 7 ,to find value of C

$\displaystyle 7 = e^{-3x^2} \times e^c$

Then ? - Apr 24th 2013, 09:10 AMtopsquarkRe: differential equation need Help
- Apr 24th 2013, 09:15 AMmastermin346Re: differential equation need Help
$\displaystyle 7 = e^{-3(0)^2} \times e^c$

$\displaystyle 7 = 1 \times e^c$

$\displaystyle C = ln 7$

right or wrong ? (Smirk) - Apr 24th 2013, 02:53 PMBobPRe: differential equation need Help
y=Aexp (-3x^2)

y (0)=7, so 7=A.

y=7exp (-3x^2). - Apr 24th 2013, 05:54 PMProve ItRe: differential equation need Help
First note that $\displaystyle \displaystyle \begin{align*} \int{\frac{1}{y}\,dy} = \ln{|y|} \end{align*}$, NOT $\displaystyle \displaystyle \begin{align*} \ln{(y)} \end{align*}$ so this means the correct working out would be

$\displaystyle \displaystyle \begin{align*} \ln{|y|} &= -3x^2 + C \\ |y| &= e^{-3x^2 + C} \\ |y| &= e^{C}\,e^{-3x^2} \\ y &= \pm e^{C}\,e^{-3x^2} \\ y &= A\,e^{-3x^2} \textrm{ where } A = \pm e^C \end{align*}$