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I am quite new to differential equations and I have to solve the following $a(t)+b(t)C(t)+s a(t)C'(t)=0$, where $s$ is some constant. I read about differential equations and at this moment my main difficulty is that $C'(t)$ is multiplied by $s a(t)$ (the examples of differential equations I have seen so far have $s a(t)=1$). Does anyone have a hint about this equation (maybe it turns out to be a "famous" equation and I am not aware about it) and how to solve it?

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  • $\begingroup$ How about dividing the equation by $sa(t)$? Then it's the same as your exemples $\endgroup$ – Tryss Jun 19 '15 at 7:40
  • $\begingroup$ Thank you for the advice. I have already did this, but I get a wrong result. I divided by $sa(t)$ and I got the following equation $s^{-1}+\frac{b(t)C(t)}{sa(t)}+C'(t)=0$. Then, assuming that $K'(t)=\frac{b(t)}{sa(t)}$ I multiplied the equation by $\exp{K(t)}$, obtaining $\exp{K(t)}\frac{b(t)C(t)}{sa(t)}+\exp{K(t)}C'(t)=\exp{K(t)}s^{-1}$. Since on the right-hand side there is a constant that multiplies $\exp K(t)$, I get a nice result, i.e. $C(t)=-a(t)/b(t)$, which is, regrettably, wrong. Any ideas? $\endgroup$ – user249255 Jun 19 '15 at 8:06
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What you did in your comment is correct up to (except a negative sign)

$$\exp{K(t)}\frac{b(t)C(t)}{sa(t)}+\exp{K(t)}C'(t)=-\exp{K(t)}s^{-1}$$

After that $$\left(\exp{K(t)}C(t)\right)'=-\exp{K(t)}s^{-1}$$

Integrate both sides $$\exp{K(t)}C(t)=-s^{-1}\int\exp{K(t)}\,dt+D\\ C(t)=-s^{-1}\exp{(-K(t))}\int\exp{K(t)}\,dt+D\exp{(-K(t))}$$ where $D$ is a constant.

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