MHB First order linear differential Equation

[Petrus]

Hello MHB,
$$(x^2+1)y'-2xy=x^2+1$$ if $$y(1)=\frac{\pi}{2}$$What I have done:
Divide evrything by $$x^2+1$$ and we got
$$y'-\frac{2xy}{x^2+1}=1$$
we got the integer factor as $$e^{^-\int\frac{2x}{x^2+1}}= e^{-ln(x^2+1)}$$
Now I get
$$(e^{-ln(x^2+1)}y)'=e^{-ln(x^2+1)}$$
and this lead me to something wrong, I am doing something wrong or?

Regards,
$$|\pi\rangle$$

 

[alyafey22]

Try simplifying the expression first , remember logarithm and exponential are (inverse) ...

 

[Petrus]

Try simplifying the expression first , remember logarithm and exponential are (inverse) ...

that means if I got right
$$e^{-ln(x^2+1)}=-x^2-1$$?

Regards,
$$|\pi\rangle$$

 

[alyafey22]

No , remember the rules of logarithm $$\log x^n = n\log x $$ .

 

[Petrus]

No , remember the rules of logarithm $$\log x^n = n\log x $$ .

Ahhh I see now! $$e^{-ln(x^2+1)}=e^{ln((x^2+1)^{-1})}= \frac{1}{x^2+1}$$
Thanks a lot Zaid! I should be able to continue!:) Thanks for fast responed and taking your time!

Regards,
$$|\pi\rangle$$