# Thread: General solution question 2

1. ## 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.

2. 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.

3. Thanks guys knew it would be simply

4. would [lny = x^2 + c] be the final answer?

5. Originally Posted by djmccabie
would [lny = x^2 + c] be the final answer?
not quite we need to simplify a bit

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

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

$y=Ae^{x^2}$

6. 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

7. 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

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