1. Differential Equation 1

Thanks!!

2. Hello, qbkr21!

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

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

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

Can you finish it now?

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

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

4. Re:

Re:

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

6. Re:

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

Thanks

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

Let's check this.

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

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

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