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First order linear differential Equation

Petrus

Well-known member
Feb 21, 2013
739
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
Jan 17, 2013
1,667
Try simplifying the expression first , remember logarithm and exponential are (inverse) ...
 

Petrus

Well-known member
Feb 21, 2013
739
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
Jan 17, 2013
1,667
No , remember the rules of logarithm \(\displaystyle \log x^n = n\log x \) .
 

Petrus

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\pi\rangle\)