Help with Differential Equations (Applied Maths)

Prodigious

Story lads,

Stuck on q3 pg 215 of Fundamental applied Maths 2nd edition

Got as far as

{Dv/(v^2+100) = {dt

How do I integrate that? It's (X)ln(v^2+100), I think, but what's x?

Thanks

clartharlear

Did you mis-type the LHS? I don't think that's the right form.
Should be 1/10 tan ^-1(v/10) ?

Prodigious

clartharlear wrote: »

Did you mis-type the LHS? I don't think that's the right form.
Should be 1/10 tan ^-1(v/10) ?

Thanks a mill!

Prodigious

The question:

A particle starts from rest at a point P and moves in a straight line subject to an acceleration = v^2 + 100 where v is velocity of the particle.
Find, correct to two decimal places, the time taken to reach 20m/s

I've got out that

V/10|= Tan10t|

But what values go where?

Thanks

clartharlear

Go back a step or two to
1/10 tan^1(V/10) = t.

From the question, you know v = 20 at the t you're looking for, so sub that in to find t.

Prodigious

Nice one, thanks for your help.

Leaving Cert Student

Can anybody tell me very quickly how I integrate a fraction such as x^2/2x + 1. Do I use substitution and set 2x + 1 = U or is it x = U?

Prodigious

Im fairly sure its the x^2 as then du/dx would be 1/2x+1 which is easily done.

Leaving Cert Student

Methememb wrote: »

Im fairly sure its the x^2 as then du/dx would be 1/2x+1 which is easily done.

sorry how does one integrate 2x+1 then? i thought the divisor has to be a single term?

Prodigious

2x + 1 becomes x^2 + x

Leaving Cert Student

Methememb wrote: »

2x + 1 becomes x^2 + x

So 1/2x+1 becomes 1/x^2+x

Prodigious

Think this is the answer, apologies for the mess.