1. ## solving ODEs

I've just started this unit at university.

A question was to solve this ODE:

y' = y

Why is it that the answer is not y= y^2/2 + c,

also should it actually be written as y(x)=ce^x, or does that not matter?

Is there an algebraic approach to do this question? I put it in wolfram, and the final two steps were:
y(x) = e^(x+c)
Simplify
=> y(x) = ce^x

I don't quite understand how that simplifies.

Any help is appreciated, thanks!

2. ## Re: solving ODEs

y'=y
You must integrate with respect to the same variable on both sides of the equation.
One integral of y' is = Integral y'(x)*dx = y(x) +c
One integral of y with respect to y is = integral y*dy = y²/2 +c
But it is of no use here, because you should integrate relatively to x , that is : Integral y*dx
Do not confuse with Integral y*dy
Since you don't know yet y(x), you cannot compute Integral y*dx. So, you need another method. The other method leads to y= c*exp(x)

3. ## Re: solving ODEs

Another example :
y(x) = x²
Integral y*dy = y²/2 +c = (x^4)/2 +c
Integral y*dx = Integral x²dx = (x^3)/3 +c
Of couse, both are different.

4. ## Re: solving ODEs

Originally Posted by JJacquelin
So, you need another method. The other method leads to y= c*exp(x)
I am still somewhat confused.
Also, are we supposed to know this 'other method' (i'm doing a second year university math subject, we use a book called Advanced engineering mathematics),
or are we supposed to be able to recognise the solution
for y'=y is y= c*exp(c)
by looking at it?

5. ## Re: solving ODEs

dy/dx = y' = y is a so called "separable" ODE. This is one of the simplest kind of ODE, as explained in all books.
Ordinary Differential Equation -- from Wolfram MathWorld (See eq. 17)
dy = y*dx
dy/y = dx
ln(y) = x +C
y = exp(x+C)
y = exp(C)*exp(x)
exp(C) = c
y = c*exp(x)

6. ## Re: solving ODEs

excellent, thank you so much!

I don't know why the textbook simply didn't show us that.

7. ## Re: solving ODEs

I have never seen a Differential Equations textbook that didn't show that! What textbook are you using?