# General solution question 2

• May 9th 2009, 10:18 AM
djmccabie
General solution question 2
Use the method of separating the variables to find the general solution of the equation

dy/dx - 2xy = 0

Could somebody please explain this method of separating variables because i have never been taught it (taught different syllabus). Only 4 mark question so guessing its nothing difficult.
• May 9th 2009, 10:19 AM
HallsofIvy
Quote:

Originally Posted by djmccabie
Use the method of separating the variables to find the general solution of the equation

dy/dx - 2xy = 0

Could somebody please explain this method of separating variables because i have never been taught it (taught different syllabus). Only 4 mark question so guessing its nothing difficult.

dy/dx= 2xy so dy/y= 2xdx. The variables are "separated". Integrate both sides.
• May 9th 2009, 10:20 AM
TheEmptySet
• May 9th 2009, 10:23 AM
djmccabie
Thanks guys knew it would be simply
• May 9th 2009, 10:26 AM
djmccabie
would [lny = x^2 + c] be the final answer?
• May 9th 2009, 10:30 AM
TheEmptySet
Quote:

Originally Posted by djmccabie
would [lny = x^2 + c] be the final answer?

not quite we need to simplify a bit

$\displaystyle \ln(y)=x^2+c \iff e^{\ln(y)}=e^{x^2+c} \iff y=e^{x^2}e^{c}$

Now we can relabel $\displaystyle e^{c}=A$ just another constant to get

$\displaystyle y=Ae^{x^2}$
• May 9th 2009, 10:45 AM
djmccabie
OK i'm struggling to understand that.

I've remembered that when theres a log in the solution I have been taught to put the constant of integration inside like this:

[lnAy = x^2] where A is the constant

so then i get

e^x² = Ay

y = (1/A)e^x²

• May 9th 2009, 11:06 AM
TheEmptySet
Quote:

Originally Posted by djmccabie
OK i'm struggling to understand that.

I've remembered that when theres a log in the solution I have been taught to put the constant of integration inside like this:

[lnAy = x^2] where A is the constant

so then i get

e^x² = Ay

y = (1/A)e^x²

$\displaystyle \frac{1}{A}=B$