# differential equation need Help

• Apr 24th 2013, 09:44 AM
mastermin346
differential equation need Help
$dy/dx = -6xy$
$\int \frac{1}{y} dy = \int -6x dx$
$ln y = -3x^2 + c$
$y = e^{-3x^2 + c}$

• Apr 24th 2013, 09:49 AM
BobP
Re: differential equation need Help
Yes, though it can be taken one step further,

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

which, subsequently, will be slightly easier to deal with.
• Apr 24th 2013, 09:55 AM
mastermin346
Re: differential equation need Help
its given y(0) = 7 ,to find value of C

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

Then ?
• Apr 24th 2013, 10:10 AM
topsquark
Re: differential equation need Help
Quote:

Originally Posted by mastermin346
its given y(0) = 7 ,to find value of C

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

Then ?

y(0) = 7 says that when x = 0 y = 7. What change does that make in the line you wrote?

-Dan
• Apr 24th 2013, 10:15 AM
mastermin346
Re: differential equation need Help
$7 = e^{-3(0)^2} \times e^c$
$7 = 1 \times e^c$
$C = ln 7$

right or wrong ? (Smirk)
• Apr 24th 2013, 03:53 PM
BobP
Re: differential equation need Help
y=Aexp (-3x^2)
y (0)=7, so 7=A.
y=7exp (-3x^2).
• Apr 24th 2013, 06:54 PM
Prove It
Re: differential equation need Help
Quote:

Originally Posted by mastermin346
$dy/dx = -6xy$
$\int \frac{1}{y} dy = \int -6x dx$
$ln y = -3x^2 + c$ You are wrong in this step
$y = e^{-3x^2 + c}$

is my answer Correct ? You need to go further

First note that \displaystyle \begin{align*} \int{\frac{1}{y}\,dy} = \ln{|y|} \end{align*}, NOT \displaystyle \begin{align*} \ln{(y)} \end{align*} so this means the correct working out would be

\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*}