# Thread: Differential Equation 1

1. ## Differential Equation 1

Thanks!!

2. Hello, qbkr21!

$\displaystyle (x^2+1)\frac{dy}{dx} \:=\:xy$
Separate the variables . . . x's on one side, y's on the other.

. . . $\displaystyle \frac{dy}{y} \:=\:\frac{x}{x^2+1}\,dx$

Then integrate: .$\displaystyle \int\frac{dy}{y} \;=\;\int\frac{x}{x^2+1}\,dx$

Can you finish it now?

3. Originally Posted by qbkr21
May I ask... what does mean the $\displaystyle dt$ over there?, we just define $\displaystyle y'=\frac{dy}{dx}.$

--

Remember the useful trick $\displaystyle \int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+k,\,\forall f(x)>0.$

4. ## Re:

Re:

5. It is faster if you define the constant as $\displaystyle \ln|c|,$ 'cause it's arbitrary.

6. ## Re:

Originally Posted by Krizalid
It is faster if you define the constant as $\displaystyle \ln|c|,$ 'cause it's arbitrary.
Example? I don't understand ...

Thanks

7. Originally Posted by Soroban
Then integrate: .$\displaystyle \int\frac{dy}{y} \;=\;\int\frac{x}{x^2+1}\,dx$
To qbkr21:

Let's check this.

Integratin' we have $\displaystyle \ln|y|=\frac12\ln(x^2+1)+c_1$

Since the constant is arbitrary, we may set it as $\displaystyle \ln|c|\implies\ln|y|=\frac12\ln(x^2+1)+\ln|c|.$

Now apply properties of logs. and you're done.