1. diff equations 2

One more diff equation. Do i still use integrating method to differentiate this one?

2. Originally Posted by moolimanj
One more diff equation. Do i still use integrating method to differentiate this one?
Your DE has a t in it ..... Did you mean:

A. $\frac{dy}{dt} = \frac{1 + y^2}{y(1 + t^2)}$

B. $\frac{dy}{dx} = \frac{1 + y^2}{y(1 + x^2)}$

C. None of the above. The DE is as posted.

3. Originally Posted by mr fantastic
Your DE has a t in it ..... Did you mean:

A. $\frac{dy}{dt} = \frac{1 + y^2}{y(1 + t^2)}$

B. $\frac{dy}{dx} = \frac{1 + y^2}{y(1 + x^2)}$

C. None of the above. The DE is as posted.
$\frac{dy}{dt} = \frac{1 + y^2}{y(1 + t^2)}$ is seperable:

$\frac{y}{1 + y^2} \, dy = \frac{1}{1 + t^2} \, dt$

$\Rightarrow \int \frac{y}{1 + y^2} \, dy = \int \frac{1}{1 + t^2} \, dt$

$\Rightarrow \frac{1}{2} \ln (1 + y^2) = \tan^{-1} (t) + C$

and you can make y the subject if you want ......

it should read dy/dt so it is A below

5. Originally Posted by moolimanj

it should read dy/dt so it is A below

6. Thanks Mr Fantastic,

just to check, if i rearrange for y I get the following:

If I then try and calculate the partucular solution that satisfies initial condition: y(1)=5 I get C =

7. Originally Posted by moolimanj
Thanks Mr Fantastic,

just to check, if i rearrange for y I get the following:

If I then try and calculate the partucular solution that satisfies initial condition: y(1)=5 I get C =
Two small mistakes I think:

1. $y = \sqrt{e^{2 (\tan^{-1} t + C)} - 1}$. The square root goes over the 1 as well.

2. It's probably easier to sub t = 1, y = 5 into $\frac{1}{2} \ln (1 + y^2) = \tan^{-1} (t) + C$:

$\frac{1}{2} \ln (1 + 25) = \tan^{-1} (1) + C$

$\Rightarrow \ln(26) = \frac{\pi}{2} + 2C$ etc.