# Math Help - solve

1. yes.. bernoullis eqn...
dy/dx +Py = Q y^x , where P and Q are functions of x alone

2. Well, technically,

dy/dx +Py = Q y^n,

but yes, it's a Bernoulli equation. Can you take it from here?

3. I think so....

4. Ok. Let me know if you get stuck.

5. sorry.. I took a break...

I got I.F as 1/x

Now answer would be

1/ux = 1/(2.x^2) + c

where u = logy

Is that correct

6. You shouldn't regard delays in replying on this forum as unusual. I have just been doing other things for a while.

I'm not sure I agree with your solution of the Bernoulli equation. Can you show your steps, please?

7. multiply the whole eqn by 1/u^2

u^-2 . du/dx + 1/(ux) = 1/x^2

put z=u^-1

now, eqn becomes

-dz/dx + z/x = x^-2

dz/dx - z/x = -x^-2

P=-1/x , Q=-x^2

IF = 1/x

solution

z . 1/x = inegral -x^-2 * 1/x dx + c

8. Ok. Its time for me to go....
Reply me when you are free
Thank you very much
love you

9. Ok, I see where you're going now. IF stands for "integrating factor", not "if" in caps. I agree with your post # 36. You should solve for u, and then substitute back in for y. y is the goal, remember.

Good work!

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