# Differential equations

1. Feb 14, 2010

### james.farrow

I have the following to problem:-

1)dy/dt = (1+y^2)/(1+4t^2) y(0)=1

This equation is seperable I arrive at

Integral 1/(1+y^2) dy = Integral 1/(1+4t^2) dt

If I let 2t=x then the RHS is in the same form as my standard integrals.

Which is Integral 1/(x^2 + a^2) = 1/a arctan(x/a)

So far so good methinks...

After a bit of 'jiggery pokery' I get

arctan(y) = arctan(2t) + C

Take tan of both sides -> y = 2t + tan(C)
Putting in initial values

1 = 2*0 + tan(C)

Therefore C = pi/4

Finally y = 2t + pi/4 -> Am I right!!

Now my second equation is thus:

(x^2 + 1)dy/dx -2xy = 2x(x^2 +1)

Putting it into standard from

dy/dx - (2xy)/(x^2 + 1) = 2x

This gives me an integrating factor 1/(x^2 +1)

Now what!?

2. Feb 14, 2010

### elect_eng

Methinks this step is an error. Note the following identity.

$${\rm \tan}(a+b)={{{\rm tan}a+{\rm tan}b}\over{1-{\rm tan}a\; {\rm tan}b}}$$

3. Feb 14, 2010

### CompuChip

However, you might want to check this step:
because tan(a + b) $\neq$ tan(a) + tan(b) in general.

4. Feb 14, 2010

### james.farrow

Cheers everyone! If I use the identity I still get C = pi/4.

What is the check??

James

5. Feb 14, 2010

### vela

Staff Emeritus
You dropped a factor of 1/2 when integrating the RHS.

6. Feb 15, 2010

### james.farrow

Can you show me how I dropped the factor of 1/2?

I have 1/(4t^2 + 1)

If x=2t then x^2=4t^2

So using 1/(x^2 + a^2) where a=1=a^2

I'm struggling with it really!!

Cheers James

7. Feb 15, 2010

### james.farrow

Ok everyone I've had a rethink on my integration...

Here goes.

Integral of 1/(1 + 4t^2) becomes

1/4 Integral of 1/(1/4 + t^2)

a^2= 1/4 so a=1/2

Standard integral 1/a arctan(x/a)

So finally after integrating I arrive at

1/2 arctan(2t)

What do you think?

James

8. Feb 15, 2010

### vela

Staff Emeritus
That's correct.

In your previous attempt, you have to replace dt by dx. Because x=2t, you had dx=2 dt, or dt=dx/2. That's where your missing factor of 1/2 went.

9. Feb 15, 2010

### james.farrow

Thanks for your help everyone! It's appreciated!

No doubt you'll be hearing from me shortly - integrating factor is next...