diff equations 2

**moolimanj** · February 14th 2008, 01:57 AM

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

**mr fantastic** · February 14th 2008, 02:23 AM

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.

**mr fantastic** · February 14th 2008, 02:27 AM

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 ......

**moolimanj** · February 14th 2008, 02:31 AM

apoligies made a mistake

it should read dy/dt so it is A below

**mr fantastic** · February 14th 2008, 02:34 AM

Originally Posted by moolimanj

apoligies made a mistake

it should read dy/dt so it is A below

OK. Read post #3.

**moolimanj** · February 14th 2008, 08:49 AM

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 =

**mr fantastic** · February 14th 2008, 01:18 PM

Originally Posted by moolimanj

Thanks Mr Fantastic,

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

Click image for larger version.

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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.

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