# A first order DE

• Mar 10th 2010, 10:29 AM
Jones
A first order DE
Hi,

I have the DE $\displaystyle \frac{dy}{dt} +\frac{dy}{dt}t^2 = e^{-y}$

If you take the reciprocal of both sides, you end up with:
$\displaystyle \frac{dt}{dy}+\frac{dt}{dy}t^{-2} = e^y$

multiply by dy:$\displaystyle dt + t^-2dt = e^y dy$

But when im then integrating this, i get
$\displaystyle t -\frac{1}{3}t^{-3} = e^y$

What's wrong here?
• Mar 10th 2010, 11:56 AM
Opalg
Quote:

Originally Posted by Jones
Hi,

I have the DE $\displaystyle \frac{dy}{dt} +\frac{dy}{dt}t^2 = e^{-y}$

If you take the reciprocal of both sides, you end up with:
$\displaystyle \frac{dt}{dy}+\frac{dt}{dy}t^{-2} = e^y$

multiply by dy:$\displaystyle dt + t^-2dt = e^y dy$

But when im then integrating this, i get
$\displaystyle t -\frac{1}{3}t^{-3} = e^y$

What's wrong here?

It is not true that the reciprocal of $\displaystyle a + b$ is $\displaystyle (1/a) + (1/b)$.
• Mar 10th 2010, 12:23 PM
Jones
Quote:

Originally Posted by Opalg
It is not true that the reciprocal of $\displaystyle a + b$ is $\displaystyle (1/a) + (1/b)$.

Right,

$\displaystyle \frac{1}{\frac{dy}{dt}+\frac{dy}{dt}t^2}$

Can you "factor out" the differential operator dy/dx, is that mathematically correct?
• Mar 10th 2010, 12:42 PM
Krizalid
since $\displaystyle y'(x)=\frac{dy}{dx}.$