First order linear differential Equation

Petrus

Well-known member
Hello MHB,
$$\displaystyle (x^2+1)y'-2xy=x^2+1$$ if $$\displaystyle y(1)=\frac{\pi}{2}$$

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

Regards,
$$\displaystyle |\pi\rangle$$

ZaidAlyafey

Well-known member
MHB Math Helper
Try simplifying the expression first , remember logarithm and exponential are (inverse) ...

Petrus

Well-known member
Try simplifying the expression first , remember logarithm and exponential are (inverse) ...
that means if I got right
$$\displaystyle e^{-ln(x^2+1)}=-x^2-1$$?

Regards,
$$\displaystyle |\pi\rangle$$

ZaidAlyafey

Well-known member
MHB Math Helper
No , remember the rules of logarithm $$\displaystyle \log x^n = n\log x$$ .

Petrus

Well-known member
No , remember the rules of logarithm $$\displaystyle \log x^n = n\log x$$ .
Ahhh I see now!!!! $$\displaystyle e^{-ln(x^2+1)}=e^{ln((x^2+1)^{-1})}= \frac{1}{x^2+1}$$
Thanks alot Zaid! I should be able to continue! Thanks for fast responed and taking your time!

Regards,
$$\displaystyle |\pi\rangle$$