learning differentual equations

[PG1995]

Hi

Please help me with this problem. I'm really stuck on this one. I'm learning to work with differential equations.

Page #1: https://img707.imageshack.us/img707/2342/page1oip.jpg
Page #2: https://img59.imageshack.us/img59/943/page2tv.jpg
Page #3: https://img228.imageshack.us/img228/7319/page3q.jpg

Link #1: https://img341.imageshack.us/img341/1465/imgch.jpg

Thanks a lot for your help.

Regards
PG

PS: I have modified the linked pages to make my queries more understandable. Thanks.

 

[MrAl]

Hi,

Not sure what your question is, but when you start to do more advanced math you'll find that the form of the equation plays a role in the solution because known solutions are often found in certain forms.

For example, we can write:
y=x+x^2+3

but often it's better to write this as:
y=x^2+x+3

where the powers of x appear in decreasing order.

That's just one very simple example, but the general idea is that you are trying to get certain variables in one place and other variables in another place, in the hopes of getting into a known solvable equation form.
With separation of variables, the idea is to get all the 'y' variables on one side and all the 'x' variables on the other side. This helps because then you can integrate both sides directly.

 

[MrAl]

Hi,

First let me ask you a question...

How are you getting:
x^2+y^2=25

with initial conditions x=0 and y=3.

How did you end up with "25" on the right side?

EDIT: My 'bad', you had y(4)=3 right, not y(0)=3.



The second part:

Following your progress in solving this...

You started with the equation:
dy/dx=y^2-4

Which is in implicit form.

You then were able to solve for 'y' explicitly ie y=f(x).

Now that you know what y is explicitly, you can find dy/dx explicitly by simple differentiation.

You will then know both 'y' and 'dy/dx' so you go back to the original equation, enter those solutions into it (which gives you an explicit form), then solve for the constant 'c'.

So you'll be working with:
f'(x)=f(x)^2-4

where f(x) is your solution for y and f'(x) is its derivative, and you'll be solving for 'c'. All we are doing here is inserting the solutions into the original equation and that gives us an equation where we can solve for 'c'. This is typical of problems like this.

The results for 'c' will be interesting. Once you simplify you'll see the answer, but if not just give a shout.