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.

Thank you for your help.


closed as off-topic by Did, Nosrati, Abcd, Namaste, Song Jan 11 at 21:56

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  • 2
    $\begingroup$ Exponentiate both sides. $\endgroup$ – Claude Leibovici Jan 11 at 9:10
  • $\begingroup$ Thank you very much! $\endgroup$ – J.Doe Jan 11 at 9:13
  • $\begingroup$ $ ln(y)=ln(1+x^2)+ln\, C \rightarrow y=C (1+x^2) $ Plug in BC $\endgroup$ – 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 $$y(-3)=10C=-1.$$


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