Differential equation $y'=2xy-x^2y'; y(-3)=1$ [closed]

Differential equation $$y'=2xy-x^2y', y(-3)=1$$

I've got the equation: $$ln(y)=ln(1+x^2)+C$$ where $$C=-ln(10)$$ and this is incorrectly.

closed as off-topic by Did, Nosrati, Abcd, amWhy, SongJan 11 at 21:56

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• Exponentiate both sides. – Claude Leibovici Jan 11 at 9:10
• Thank you very much! – J.Doe Jan 11 at 9:13
• $ln(y)=ln(1+x^2)+ln\, C \rightarrow y=C (1+x^2)$ Plug in BC – Narasimham Jan 11 at 18:17

Rewrite your equation as $$\frac{dy}{dx}=\frac{2xy}{1+x^2}.$$ The general solution of $$\frac{dy}{dx}=f(x)y$$ is $$y=Ce^{\int f(x)dx}$$ where $$C$$ is an arbitrary constant. Here you have $$f(x)=\frac{2x}{1+x^2}$$ so the solution is $$y=Ce^{\ln(1+x^2)}=C(1+x^2)$$ and $$y(-3)=10C=-1.$$