Question

I'm having a little trouble with this ODE. (1+x^2)y'-2xy=2x(1+x^2).
p(x) should equal to -2x/(1+x^2)
Therefor the integrating factor should be e^-ln(1+x^2)=-(1+x^2)
Then -(1+x^2)y=integral(-2x(x^2 + 1))
meaning -x^4/-2(x^2 +1) - (x^2/-(x^2 +1)) +c/(x^2 +1). WolframAlpha shows that the answer is "y(x) = c_1 (x^2+1)+(x^2+1) ln(x^2+1)". What am I doing wrong?

Answer 1

\[(1+x^2)y'-2xy=2x(1+x^2)\enspace/:(1+x^2)^2\]
\[\frac{(1+x^2)y'-2xy}{(1+x^2)^2}=\frac{2x}{1+x^2}\]
\[\left(\frac{y}{1+x^2}\right)'=\left(\ln{(1+x^2)}\right)'\]
\[\frac{y}{1+x^2}=\ln{C_1\,(1+x^2)}\]
\[y=(1+x^2)\cdot\ln{C_1\,(1+x^2)}\,\,,\,\,C_1>0\]
or
\[y=C_2(1+x^2)+(1+x^2)\cdot\ln{(1+x^2)}\,\,,C_2\in\mathbb{R}\]

Answer 2

\[C_2=\ln{C_1}\]

Answer 3

But if you divide the entire equation by \[(1+x^2)^2\] then it won't be in the correct form for a linear equation?